# How do you find the zeros, real and imaginary, of  y = 75x^2 +100x + 23  using the quadratic formula?

Jun 4, 2017

$x = - \frac{2}{3} \pm \frac{\sqrt{31}}{15}$

#### Explanation:

Given: $y = 75 {x}^{2} + 100 x + 23$

Using the quadratic formula requires the equation to be in the form $A {x}^{2} + B x + C = 0$

Quadratic formula: $x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$

For the given, $A = 75 , B = 100 , C = 23$:

$x = \frac{- 100 \pm \sqrt{10 , 000 - 4 \cdot 75 \cdot 23}}{2 \cdot 75}$

$x = \frac{- 100 \pm \sqrt{10 , 000 - 6900}}{150} = \frac{- 100 \pm \sqrt{3100}}{150}$

$x = - \frac{10}{15} \pm \frac{\sqrt{25 \cdot 4 \cdot 31}}{150} = - \frac{10}{15} \pm \frac{\sqrt{25} \sqrt{4} \sqrt{31}}{150}$

$x = - \frac{2}{3} \pm \frac{10 \sqrt{31}}{150} = - \frac{2}{3} \pm \frac{\sqrt{31}}{15}$