# How do you find the roots, real and imaginary, of y=-15^2 + 12x +34  using the quadratic formula?

Jul 19, 2017

See a solution process below:

(Assuming the problem is: $y = - 15 {x}^{2} + 12 x + 34$)

#### Explanation:

For $y = a {x}^{2} + b x + c = 0$, the values of $x$ which are the roots to the equation are found by:

$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$

Substituting $- 15$ for $a$; $12$ for $b$ and $34$ for $c$ gives:

$x = \frac{- 12 \pm \sqrt{{12}^{2} - \left(4 \cdot - 15 \cdot 34\right)}}{2 \cdot - 15}$

$x = \frac{- 12 \pm \sqrt{144 - \left(- 2040\right)}}{- 30}$

$x = \frac{- 12 \pm \sqrt{144 + 2040}}{- 30}$

$x = \frac{- 12 \pm \sqrt{2184}}{- 30}$

$x = \frac{- 12}{- 30} \pm \frac{\sqrt{2184}}{- 30}$

$x = \frac{2}{5} \pm \frac{\sqrt{2184}}{- 30}$

$x = \frac{2}{5} \pm \frac{\sqrt{2184}}{- 30}$

$x = \frac{2}{5} \pm \frac{\sqrt{4 \cdot 546}}{- 30}$

$x = \frac{2}{5} \pm \frac{\sqrt{4} \sqrt{546}}{- 30}$

$x = \frac{2}{5} \pm \frac{2 \sqrt{546}}{- 30}$

$x = \frac{2}{5} \pm \frac{\sqrt{546}}{- 15}$

$x = \frac{2}{5} - \frac{\sqrt{546}}{15}$ and $x = \frac{2}{5} + \frac{\sqrt{546}}{15}$