# How do you find the zeros, real and imaginary, of y=x^2-8x-3 using the quadratic formula?

Jun 9, 2016

This function hs $2$ real zeros:

${x}_{1} = 4 - \sqrt{19}$

${x}_{2} = 4 + \sqrt{19}$

#### Explanation:

To calculate the zeros of a quadratic function first you calculate the determinant:

$y = {x}^{2} - 8 x - 3$

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

$\sqrt{\Delta} = \sqrt{76} = \sqrt{4 \times 19} = 2 \sqrt{19}$

$\Delta \ge 0$, so the function has $2$ real zeros. (If $\Delta < 0$ then the zeros would be complex numbers)

${x}_{1} = \frac{- b - \sqrt{\Delta}}{2 a}$

${x}_{1} = \frac{8 - 2 \sqrt{19}}{2} = 4 - \sqrt{19}$

${x}_{2} = \frac{- b + \sqrt{\Delta}}{2 a}$

${x}_{1} = \frac{8 + 2 \sqrt{19}}{2} = 4 + \sqrt{19}$