How do you find the roots, real and imaginary, of y=-5(x-3)^2-(x-3)^2-10  using the quadratic formula?

Apr 26, 2016

$3 - i 10 \sqrt{2}$ and $3 + i 10 \sqrt{2}$ are roots of given equation.

Explanation:

For equation $a {x}^{2} + b x + c = 0$, quadratic formula gives $x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$.

In equation $y = - 5 {\left(x - 3\right)}^{2} - {\left(x - 3\right)}^{2} - 10 = - 6 {\left(x - 3\right)}^{2} - 10$. Hence, $\left(x - 3\right) = \frac{0 \pm \sqrt{{0}^{2} - 4 \left(- 5\right) \left(- 10\right)}}{2 \times \left(- 5\right)} = \frac{\pm \sqrt{- 200}}{- 10} = \pm i 10 \frac{\sqrt{2}}{- 10} = \pm i 10 \sqrt{2}$

Hence, $x = 3 \pm i 10 \sqrt{2}$ and these are roots of given equation.