Another way of solving a quadratic equation on the form of

$\\ ax^{2}+bx+c=0 \\$

Is to used the quadratic formula. It tells us that the solutions of the quadratic equation are

$\\ x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \\\\\: \: where\: \: a\neq 0\: \: and\: \: b^{2}-4ac\geq 0 \\$

Example:

Solve the equation

$\\ x^{2}{\color{green} \, -\, 3}x{\color{blue} \, -\, 10}=0 \\\\ x=\frac{-\left ( {\color{green} -\, 3} \right )\pm \sqrt{\left ( {\color{green} -\, 3} \right )^{2}-4\cdot \left ( 1\cdot {\color{blue} -\, 10} \right )}}{1\cdot 2} \\\\ x=\frac{3\pm \sqrt{9+40}}{2} \\\\\left\{\begin{matrix} {\color{red} x_{1}}= &\frac{3{\color{red} \, +\, }\sqrt{49}}{2}= & \frac{3+7}{2}= & \frac{10}{2}= &{\color{red} 5} \\ & & & &\\ {\color{red} x_{2}}= & \frac{3{\color{red} \, -\, }\sqrt{49}}{2}= & \frac{3-7}{2}=& \frac{-4}{2}= & {\color{red} -2} \end{matrix}\right. \\$

The expression

$\\ b^{2}-4ac \\$

Within the quadratic formula is called the discriminant. The discriminant can be used to determine how many solutions the quadratic equation has.

$\\ \begin{matrix} if\: \: b^{2}-4ac>0 & &\: \: \: \: \: \: \: \: \: \: 2\: \: solutions \\ if\: \: b^{2}-4ac=0 & & \: \: \: \: \: \: \: \: \: \: \: \: 1\: \: solution\\ if\: \: b^{2}-4ac<0 & & no\: \: real\: \: solution \end{matrix} \\$

Here you can check that you've got the right solution

Video lesson: Solve the equation using the quadratic formula

$\\x^{2} - 3x-10 \\$