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

Dec 7, 2015

Plug the equation into the formula and solve for x. These will be your zeroes.

#### Explanation:

$y = a {x}^{2} + b x + c$
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
plug in the values from the equation
$x = \frac{- 3 \pm \sqrt{{3}^{2} - 4 \left(1\right) \left(7\right)}}{2 \left(1\right)}$
then solve for x...
$x = - \frac{3}{2} \pm \frac{\sqrt{- 19}}{2}$

$x = \frac{- 3 \pm i \sqrt{19}}{2}$
This gives 2 imaginary zeroes, but does not give any real zeroes, and if you graph the equation you will see that the equation does not have any.