How to Solve Quadratic Equations: 3 Methods with Worked Examples

A quadratic equation is any equation that can be written in the form

$$a{x}^2+bx+c=0,a\ne 0$$ax^2 + bx + c = 0, a ≠ 0

where $a$a, $b$b and $c$c are numbers. The highest power of $x$x is 2, so a quadratic has at most two solutions (also called roots). Graphically, the solutions are the points where the parabola $y=a{x}^2+bx+c$y = ax^2 + bx + c crosses the $x$x-axis.

There are three standard ways to solve quadratic equations: factorising, completing the square and the quadratic formula. This guide works through each one, shows how to choose between them in an exam, and ends with practice questions and answers. You can try each method on your own equation as you read:

How to solve quadratic equations: start with the standard form

Every method begins the same way: move every term to one side so the equation equals zero. If a question gives you

$$3x^2=5x+2$$3x^2 = 5x + 2

subtract $5x$5x and $2$2 from both sides first:

$$3x^2-5x-2=0$$3x^2 - 5x - 2 = 0

Now you can read off $a=3$a = 3, $b=-5$b = -5 and $c=-2$c = -2. Keep the signs with the numbers — most mistakes in quadratics are sign mistakes.

Method 1: solve by factorising

Factorising is the fastest method when the roots are whole numbers or simple fractions. The idea is to write the quadratic as a product of two brackets. If a product is zero, one of its factors must be zero, so each bracket gives one solution.

Example 1: solve $x^2-x-12=0$x^2 - x - 12 = 0.

We need two numbers that multiply to $-12$-12 and add to $-1$-1. They are $-4$-4 and $3$3:

$$(x-4)(x+3)=0\Rightarrow x=4\text{ or }x=-3$$(x - 4)(x + 3) = 0 ⇒ x = 4 or x = -3

Example 2 (when a is not 1): solve $2x^2-7x+3=0$2x^2 - 7x + 3 = 0 by splitting the middle term.

1. Multiply $a$a and $c$c: $2\times 3=6$2 × 3 = 6.
2. Find two numbers that multiply to 6 and add to −7: they are −6 and −1.
3. Split the middle term: $2x^2-6x-x+3=0$2x^2 - 6x - x + 3 = 0.
4. Group: $2x(x-3)-1(x-3)=0\Rightarrow (2x-1)(x-3)=0$2x(x - 3) - 1(x - 3) = 0 ⇒ (2x - 1)(x - 3) = 0.
5. So $x=\frac{1}{2}$x = 1/2 or $x=3$x = 3.

The same steps solve the standard-form example above: for $3x^2-5x-2=0$3x^2 - 5x - 2 = 0, the product $ac$ac is $-6$-6, the numbers are $-6$-6 and $1$1, and the equation factorises to $(3x+1)(x-2)=0$(3x + 1)(x - 2) = 0, so $x=2$x = 2 or $x=-\frac{1}{3}$x = -1/3.

Two special cases come up so often that they are worth recognising on sight:

• No constant term. For $x^2-5x=0$x^2 - 5x = 0, take out the common factor: $x(x-5)=0$x(x - 5) = 0, so $x=0$x = 0 or $x=5$x = 5. Do not divide both sides by $x$x — that throws away the solution $x=0$x = 0.
• Difference of two squares. For $x^2-9=0$x^2 - 9 = 0, use $x^2-9=(x-3)(x+3)$x^2 - 9 = (x - 3)(x + 3), so $x=\pm 3$x = ± 3.

If you cannot find the two numbers within about thirty seconds, the roots are probably not whole numbers. Switch to one of the next two methods rather than guessing.

Method 2: solve by completing the square

Completing the square rewrites the quadratic so that $x$x appears only once, inside a squared bracket. It always works, and it also gives you the vertex (turning point) of the parabola, which some questions ask for.

Example 3: solve $x^2+6x+2=0$x^2 + 6x + 2 = 0.

Move the constant across, then add the square of half the coefficient of $x$x to both sides. Half of 6 is 3, and $3^2=9$3^2 = 9:

$$x^2+6x=-2\Rightarrow x^2+6x+9=7\Rightarrow (x+3{)}^2=7$$x^2 + 6x = -2 ⇒ x^2 + 6x + 9 = 7 ⇒ (x + 3)^2 = 7

Take the square root of both sides — remembering both signs — and subtract 3:

$$x=-3\pm \sqrt{7}$$x = -3 ± √(7)

Written as $y=(x+3{)}^2-7$y = (x + 3)^2 - 7, the same work shows that the parabola has its lowest point at $(-3,-7)$(-3, -7).

Example 4 (when a is not 1): solve $2x^2+8x-10=0$2x^2 + 8x - 10 = 0.

Divide every term by $a=2$a = 2 first so the $x^2$x^2 term has coefficient 1:

$$x^2+4x-5=0\Rightarrow x^2+4x+4=9\Rightarrow (x+2{)}^2=9$$x^2 + 4x - 5 = 0 ⇒ x^2 + 4x + 4 = 9 ⇒ (x + 2)^2 = 9

So $x+2=\pm 3$x + 2 = ± 3, which gives $x=1$x = 1 or $x=-5$x = -5.

Method 3: solve with the quadratic formula

The quadratic formula solves every quadratic equation, including ones whose roots are irrational or complex. In India it is taught as the Sridharacharya formula:

$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$$x = frac -b ± sqrt b^2 - 4ac2a

It is not magic: it is what you get when you complete the square on the general equation $a{x}^2+bx+c=0$ax^2 + bx + c = 0. Dividing by $a$a and completing the square gives

$${\left(x+\frac{b}{2a}\right)}^2=\frac{b^2-4ac}{4a^2}$$(x + b/2a)^2 = frac b^2 - 4ac4a^2

and taking square roots and subtracting $\frac{b}{2a}$b/2a produces the formula.

Example 5: solve $3x^2-5x-1=0$3x^2 - 5x - 1 = 0. Here $a=3$a = 3, $b=-5$b = -5, $c=-1$c = -1.

$$x=\frac{-(-5)\pm \sqrt{(-5{)}^2-4(3)(-1)}}{2(3)}=\frac{5\pm \sqrt{25+12}}{6}=\frac{5\pm \sqrt{37}}{6}$$x = frac -(-5) ± sqrt (-5)^2 - 4(3)(-1)2(3) = (5 ± √(25 + 12))/(6) = (5 ± √(37))/(6)

Since $\sqrt{37}\approx 6.083$√(37) ≈ 6.083, the solutions are $x\approx 1.847$x ≈ 1.847 and $x\approx -0.180$x ≈ -0.180. This equation does not factorise nicely, which is exactly when the formula earns its place.

The discriminant: how many solutions?

The expression under the square root, $\Delta =b^2-4ac$Δ = b^2 - 4ac, is called the discriminant. Work it out first and you know what kind of answer to expect:

For the last example the formula still works if you allow complex numbers:

$$x=\frac{-4\pm \sqrt{-4}}{2}=-2\pm i$$x = (-4 ± √(-4))/(2) = -2 ± i

If the discriminant is a perfect square (0, 1, 4, 9, 16, …), the roots are rational — a strong hint that the quadratic factorises.

Common mistakes when solving quadratic equations

• Not setting the equation to zero. Factorising $x^2-5x=6$x^2 - 5x = 6 as $x(x-5)=6$x(x - 5) = 6 tells you nothing; rearrange to $x^2-5x-6=0$x^2 - 5x - 6 = 0 first.
• Dividing by x. It loses the root $x=0$x = 0.
• Dropping the minus in −b. When $b=-5$b = -5, the formula starts with $-b=5$-b = 5.
• Squaring a negative incorrectly. $(-5{)}^2=25$(-5)^2 = 25. On a calculator, type the brackets: without them, $-5^2$-5^2 is read as $-25$-25.
• Dividing only part of the numerator by 2a. The whole of $-b\pm \sqrt{b^2-4ac}$-b ± sqrt b^2 - 4ac is divided by $2a$2a.
• Forgetting the ± sign. When you take a square root while completing the square, there are two possibilities.

Check your solutions

The quickest check is to substitute each root back into the original equation. For $x=3$x = 3 in $2x^2-7x+3=0$2x^2 - 7x + 3 = 0: $2(9)-21+3=0$2(9) - 21 + 3 = 0 ✓.

For irrational roots, use the sum and product of the roots instead. For any quadratic,

$$x_1+x_2=-\frac{b}{a},x_1{x}_2=\frac{c}{a}$$x_1 + x_2 = -b/a, x_1 x_2 = c/a

For Example 5 the two roots add to $\frac{10}{6}=\frac{5}{3}$10/6 = 5/3 and multiply to $\frac{25-37}{36}=-\frac{1}{3}$(25 - 37)/(36) = -1/3, which match $-\frac{b}{a}$-b/a and $\frac{c}{a}$c/a exactly.

Which method should I use?

In an exam, a good habit is to work out the discriminant first. A perfect square means try factorising; anything else means go straight to the formula.

Practice questions

Try these, then check your answers below.

1. $x^2+5x+6=0$x^2 + 5x + 6 = 0
2. $4x^2-4x+1=0$4x^2 - 4x + 1 = 0
3. $x^2-2x-2=0$x^2 - 2x - 2 = 0
4. $2x^2+3x+5=0$2x^2 + 3x + 5 = 0
5. $6x^2-x-2=0$6x^2 - x - 2 = 0

Answers:

1. $(x+2)(x+3)=0$(x + 2)(x + 3) = 0, so $x=-2$x = -2 or $x=-3$x = -3.
2. $(2x-1{)}^2=0$(2x - 1)^2 = 0, so $x=\frac{1}{2}$x = 1/2 (a repeated root; $\Delta =0$Δ = 0).
3. $\Delta =12$Δ = 12, so $x=1\pm \sqrt{3}$x = 1 ± √(3).
4. $\Delta =9-40=-31$Δ = 9 - 40 = -31: no real roots; $x=\frac{-3\pm i\sqrt{31}}{4}$x = (-3 ± i√(31))/(4).
5. $(2x+1)(3x-2)=0$(2x + 1)(3x - 2) = 0, so $x=-\frac{1}{2}$x = -1/2 or $x=\frac{2}{3}$x = 2/3.

FAQ: solving quadratic equations

Can a quadratic equation have only one solution? Yes. When the discriminant is zero, both roots are the same number. The parabola just touches the $x$x-axis at its vertex.

What if a = 0? Then the equation is not quadratic. It is the linear equation $bx+c=0$bx + c = 0, with the single solution $x=-\frac{c}{b}$x = -c/b.

Is the quadratic formula the same as the Sridharacharya formula? Yes — they are two names for the same formula. CBSE textbooks use Sridharacharya's name; GCSE, A-level and AP courses call it the quadratic formula.

Which method is quickest? Factorising, when it works. The formula is slower but never fails, so it is the safe choice when you are short of time.

To practise how to solve quadratic equations with instant feedback, type any equation into the quadratic formula calculator. It shows all three methods side by side and checks every root by substitution.