# How do you find the zeros, real and imaginary, of  y=3x^2-7x-5  using the quadratic formula?

Sep 14, 2016

Zeros are real and are $\frac{7 - \sqrt{109}}{6}$ and $\frac{7 + \sqrt{109}}{6}$

#### Explanation:

The zeros of $y = a {x}^{2} + b x + c$ are given by quadratic formula

and are $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Hence, zeros of $y = 3 {x}^{2} - 7 x - 5$ are given by

$\frac{- \left(- 7\right) \pm \sqrt{{\left(- 7\right)}^{2} - 4 \times 3 \times \left(- 5\right)}}{2 \times 3}$

= $\frac{7 \pm \sqrt{49 + 60}}{6}$

= $\frac{7 \pm \sqrt{109}}{6}$

Hence zeros are real and are $\frac{7 - \sqrt{109}}{6}$ and $\frac{7 + \sqrt{109}}{6}$