The Quadratic Formula

When no pair can be found

Factoring solves $x^2-2x-3=0$x² − 2x − 3 = 0 quickly: the numbers −3 and 1 multiply to −3 and add to −2, so (x − 3)(x + 1) = 0, and x = 3 or x = −1.

Now try $x^2+4x+1=0$x² + 4x + 1 = 0. The only whole-number pairs that multiply to 1 are 1 and 1, with sum 2, and −1 and −1, with sum −2. Neither adds to 4, so the left side does not factor with whole numbers. The equation still has solutions, but they are not whole numbers. The quadratic formula finds the solutions of any quadratic equation, whether it factors or not.

Reading a, b and c

First write the equation in the form $a{x}^2+bx+c=0$ax² + bx + c = 0, with every term on one side and 0 on the other. Then a is the coefficient of $x^2$x², b is the coefficient of x, and c is the constant term.

Each number keeps its sign. In $x^2-2x-3=0$x² − 2x − 3 = 0, a = 1, b = −2 and c = −3. When $x^2$x² has no number in front of it, a = 1. When there is no x term, b = 0.

An equation that is not yet in this form must be rearranged first. For $3x^2=5-2x$3x² = 5 − 2x, add 2x to both sides and subtract 5 from both sides: $3x^2+2x-5=0$3x² + 2x − 5 = 0. Now a = 3, b = 2 and c = −5.

The formula

The solutions of $a{x}^2+bx+c=0$ax² + bx + c = 0 are $x=\left(-b\pm \sqrt{b^2-4ac}\right)$x = (−b ± √(b² − 4ac)) / 2a.

The sign $\pm$± is read "plus or minus". It stands for two calculations: one where the square root is added to −b, and one where it is subtracted. That is why the formula gives both solutions at once.

The line of the fraction runs under the whole of $-b\pm \sqrt{b^2-4ac}$−b ± √(b² − 4ac). All of it is divided by 2a, not only the square root.

Using it

Solve $x^2-2x-3=0$x² − 2x − 3 = 0 with the formula. Here a = 1, b = −2 and c = −3.

Work out $b^2-4ac$b² − 4ac first, on its own: ${\left(-2\right)}^2-4\times 1\times \left(-3\right)=4+12=16$(−2)² − 4 × 1 × (−3) = 4 + 12 = 16. Its square root is 4. Next, −b is −(−2) = 2, and 2a is 2. So $x=\left(2\pm 4\right)$x = (2 ± 4) / 2.

The plus gives (2 + 4) / 2 = 6 / 2 = 3, and the minus gives (2 − 4) / 2 = −2 / 2 = −1. These are 3 and −1, the same solutions that factoring gave.

The working for $x^2-2x-3=0$x² − 2x − 3 = 0, one quantity at a time. The last row is $x=\left(2\pm 4\right)$x = (2 ± 4) / 2, which is 3 or −1.

The solutions on the graph

Draw the graph of $y=x^2-2x-3$y = x² − 2x − 3 by working out y for each x. The graph of a quadratic is a curve called a parabola. The solutions of $x^2-2x-3=0$x² − 2x − 3 = 0 are the values of x that make y equal to 0, and y is 0 exactly where the curve meets the x-axis. So the curve crosses the x-axis at x = −1 and x = 3.

The curve $y=x^2-2x-3$y = x² − 2x − 3 crosses the x-axis at x = −1 and x = 3, the two solutions.

Solutions with a square root in them

Return to $x^2+4x+1=0$x² + 4x + 1 = 0. Here a = 1, b = 4 and c = 1, so $b^2-4ac=16-4=12$b² − 4ac = 16 − 4 = 12, and $x=\left(-4\pm \sqrt{12}\right)$x = (−4 ± √12) / 2.

Simplify the radical: $\sqrt{12}=\sqrt{4}\times \sqrt{3}=2\sqrt{3}$√12 = √4 × √3 = 2√3. Then $x=\left(-4\pm 2\sqrt{3}\right)$x = (−4 ± 2√3) / $2=-2\pm \sqrt{3}$2 = −2 ± √3. The two exact solutions are $-2+\sqrt{3}$−2 + √3 and $-2-\sqrt{3}$−2 − √3, which are about −0.27 and −3.73.

When a is not 1, 2a is not 2. For $2x^2+3x-4=0$2x² + 3x − 4 = 0, a = 2, b = 3 and c = −4, so $b^2-4ac=9+32=41$b² − 4ac = 9 + 32 = 41 and $x=\left(-3\pm \sqrt{41}\right)$x = (−3 ± √41) / 4. Since $\sqrt{41}$√41 is about 6.403, the solutions are about 0.85 and −2.35.

The curve $y=x^2+4x+1$y = x² + 4x + 1 crosses the x-axis between whole numbers, at $-2-\sqrt{3}$−2 − √3 and $-2+\sqrt{3}$−2 + √3, which are about −3.7 and −0.3.

When $b^2-4ac$b² − 4ac is negative

For $x^2+2x+5=0$x² + 2x + 5 = 0, $b^2-4ac=4-20=-16$b² − 4ac = 4 − 20 = −16. No real number squares to a negative number, so $\sqrt{-16}$√(−16) is not a real number, and the equation has no real solutions. Its curve never reaches the x-axis. The value of $b^2-4ac$b² − 4ac tells you how many solutions to expect before you finish the working.

The usual mistakes

Dividing only the square root by 2a. In $x=\left(2\pm 4\right)$x = (2 ± 4) / 2, the 2 and the 4 are both divided by 2, giving 3 and −1, not $2\pm 2$2 ± 2.

Losing a sign. When b = −2, −b is +2, and $b^2$b² is ${\left(-2\right)}^2=4$(−2)² = 4, not −4. When c is negative, −4ac is positive: −4 × 1 × (−3) = +12.

Using only the plus. The $\pm$± stands for two calculations, and a quadratic with $b^2-4ac$b² − 4ac above zero has two solutions.

An equation from an area

In an application, the equation has to be written first. A rectangle is 2 cm longer than it is wide, and its area is 10 cm². Let the width be x cm. The length is then (x + 2) cm, and the area gives x(x + 2) = 10. Expand and subtract 10 from both sides: $x^2+2x-10=0$x² + 2x − 10 = 0.

No two whole numbers multiply to −10 and add to 2, so use the formula with a = 1, b = 2 and c = −10. $b^2-4ac=4+40=44$b² − 4ac = 4 + 40 = 44, and $\sqrt{44}=2\sqrt{11}$√44 = 2√11, so $x=\left(-2\pm 2\sqrt{11}\right)$x = (−2 ± 2√11) / $2=-1\pm \sqrt{11}$2 = −1 ± √11. That is about 2.32 or −4.32. A width cannot be negative, so the width is about 2.32 cm.