# How do you find the zeros, real and imaginary, of y=5x^2-x+12 using the quadratic formula?

Jan 9, 2017

The equation has $2$ complex zeros. See explanation.

#### Explanation:

First we have to calculate the discriminant:

## $\Delta = {b}^{2} - 4 a c$

$\Delta = {\left(- 1\right)}^{2} - 4 \times 5 \times 12 = 1 - 240 = - 239$

The discriominant is negative, so the equation has 2 complex zeros:

${x}_{1} = \frac{- b - \sqrt{\Delta}}{2 a} = \frac{1 - \sqrt{- 239}}{10} = 0.1 - 0.1 \cdot \sqrt{239} i$

${x}_{2} = \frac{- b + \sqrt{\Delta}}{2 a} = \frac{1 + \sqrt{- 239}}{10} = 0.1 + 0.1 \cdot \sqrt{239} i$