# How do you find the solution to the quadratic equation x^2 + 8x - 3 = 0?

May 14, 2015

${x}^{2} + 8 x - 3 = 0$ is of the form $a {x}^{2} + b x + c = 0$ with $a = 1$, $b = 8$ and $c = - 3$.

The discriminant:

$\Delta = {b}^{2} - 4 a c = {8}^{2} - 4 \cdot 1 \cdot \left(- 3\right) = 64 + 12 = 76$.

Since $\Delta > 0$, the quadratic equation has 2 distinct real solutions. It is not a perfect square, so those solutions will not be integers or rational numbers.

The general formula to solve a quadratic is:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} = \frac{- b \pm \sqrt{\Delta}}{2 a}$

In our case:

$x = \frac{- 8 \pm \sqrt{76}}{2} = \frac{- 8 + = \sqrt{4 \cdot 19}}{2} = \frac{- 8 \pm 2 \sqrt{19}}{2}$

$= - 4 \pm \sqrt{19}$