# How do you find the zeros, real and imaginary, of  y = 7x^2 +12x + 2  using the quadratic formula?

Jan 17, 2017

The solutions are $S = \left\{\frac{- 6 + \sqrt{22}}{7} , \frac{- 6 - \sqrt{22}}{7}\right\}$

#### Explanation:

We compare this equation

$7 {x}^{2} + 12 x + 2 = 0$

to

$a {x}^{2} + b + c = 0$

We start by calculating the discriminant

$\Delta = {b}^{2} - 4 a c = {12}^{2} - 4 \cdot 7 \cdot 2 = 144 - 56 = 88$

Therefore,

As $\Delta > 0$, there are 2 real roots

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

${x}_{1} = \frac{- 12 + \sqrt{88}}{14} = \frac{- 6 + \sqrt{22}}{7}$

${x}_{2} = \frac{- 12 - \sqrt{88}}{14} = \frac{- 6 - \sqrt{22}}{7}$