The roots of the quadratic equation

 $ax^{2}+bx+c=0\qquad{a,b,c\in\mathbbmss{R},a\neq 0}$

are given by the formula

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

The number $\Delta=b^{2}-4ac$ is called the discriminant of the equation. If $\Delta>0$, there are two different real roots, if $\Delta=0$ there is a single real root, and if $\Delta<0$ there are no real roots (but two different complex roots).

Let’s work a few examples.

First, consider $2x^{2}-14x+24=0$. Here $a=2$, $b=-14$, and $c=24$. Substituting in the formula gives us

 $x=\frac{14\pm\sqrt{(-14)^{2}-4\cdot 2\cdot 24}}{2\cdot 2}=\frac{14\pm\sqrt{4}}% {4}=\frac{14\pm 2}{4}=\frac{7\pm 1}{2}.$

So we have two solutions (depending on whether we take the sign $+$ or $-$): $x=\frac{8}{2}=4$ and $x=\frac{6}{2}=3$.

Now we will solve $x^{2}-x-1=0$. Here $a=1$, $b=-1$, and $c=-1$, so

 $x=\frac{1\pm\sqrt{(-1)^{2}-4(1)(-1)}}{2}=\frac{1\pm{\sqrt{5}}}{2},$

and the solutions are $x=\frac{1+\sqrt{5}}{2}$ and $x=\frac{1-\sqrt{5}}{2}$.