# How do you find the roots, real and imaginary, of y= 5x^2-6x +35  using the quadratic formula?

Apr 9, 2017

$x = \frac{3}{5} \pm \frac{\sqrt{166} i}{5}$

#### Explanation:

Use the quadratic formula $x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A}$ when the quadratic equation is in the form $A {x}^{2} + B x + C = 0$

For $y = 5 {x}^{2} - 6 x + 35 = 0 , \text{ } A = 5 , B = - 6 , C = 35$

$x = \frac{- B \pm \sqrt{{B}^{2} - 4 A C}}{2 A} = \frac{6 \pm \sqrt{36 - 4 \left(5\right) \left(35\right)}}{2 \cdot 5}$

$x = \frac{6 \pm \sqrt{- 664}}{10} = \frac{3}{5} \pm \frac{\sqrt{- 4 \cdot 166}}{10} = \frac{3}{5} \pm \frac{2 \sqrt{166} i}{10}$

$x = \frac{3}{5} \pm \frac{\sqrt{166} i}{5}$