# How do you find the zeros, real and imaginary, of y=-5x^2-x+5 using the quadratic formula?

Nov 23, 2015

This equation has 2 real zeros:

${x}_{1} = \frac{- 1 + \sqrt{101}}{10}$

${x}_{2} = \frac{- 1 - \sqrt{101}}{10}$

#### Explanation:

To find zeros of quadratic equation you have to calculate $\Delta$ first:

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \cdot \left(- 5\right) \cdot 5 = 1 + 100 = 101$

$\Delta > 0$ so the formula has 2 real zeros.

${x}_{1} = \frac{- b - \sqrt{\Delta}}{2 a} = \frac{1 - \sqrt{101}}{- 10} = \frac{- 1 + \sqrt{101}}{10}$

${x}_{2} = \frac{- b + \sqrt{\Delta}}{2 a} = \frac{1 + \sqrt{101}}{- 10} = \frac{- 1 - \sqrt{101}}{10}$

If $\Delta < 0$ then there would be 2 imaginary zeros (2 complex conjugate numbers)

If $\Delta = 0$ there would be a single real zero