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

Dec 28, 2016

We have real zeros $\frac{- 3 + \sqrt{73}}{16}$ and $\frac{- 3 - \sqrt{73}}{16}$

#### Explanation:

Quadratic formula gives the zeros of a quadratic function $y = a {x}^{2} + b x + c$ as $\frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

Hence, zeros of a function $y = 8 {x}^{2} + 3 x - 2$ are

$\frac{- 3 \pm \sqrt{{3}^{2} - 4 \times 8 \times \left(- 2\right)}}{2 \times 8}$

or $\frac{- 3 \pm \sqrt{9 + 64}}{16}$

or $\frac{- 3 \pm \sqrt{73}}{16}$

i.e. $\frac{- 3 + \sqrt{73}}{16}$ and $\frac{- 3 - \sqrt{73}}{16}$

i.e. we have real zeros.