# How do you find the roots, real and imaginary, of y=-14 x^2 +18x +16-(x-3)^2  using the quadratic formula?

Aug 6, 2016

$x \in \left\{\frac{4}{5} - \frac{\sqrt{249}}{15} , \frac{4}{5} + \frac{\sqrt{249}}{15}\right\}$

#### Explanation:

The quadratic formula gives us the solutions to a quadratic equation of the form $a {x}^{2} + b x + c = 0$. Thus, to use it, we must work the equation into that form.

$- 14 {x}^{2} + 18 x + 16 - {\left(x - 3\right)}^{2}$

$= - 14 {x}^{2} + 18 x + 16 - \left({x}^{2} - 6 x + 9\right)$

$= - 14 {x}^{2} + 18 x + 16 - {x}^{2} + 6 x - 9$

$= - 15 {x}^{2} + 24 x + 7$

Thus, we can now look for the roots as the solutions to $a {x}^{2} + b x + c = 0$, with $a = - 15 , b = 24 , c = 7$

Plugging our values into the formula, we get

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

$= \frac{- 24 \pm \sqrt{{24}^{2} - 4 \left(- 15\right) \left(7\right)}}{2 \left(- 15\right)}$

$= - \frac{- 24 \pm \sqrt{576 + 420}}{30}$

$= \frac{24 \pm \sqrt{996}}{30}$

$= \frac{4}{5} \pm \frac{\sqrt{249}}{15}$

Thus, the roots of the given equation are $\left\{\frac{4}{5} - \frac{\sqrt{249}}{15} , \frac{4}{5} + \frac{\sqrt{249}}{15}\right\}$