# How do you find the roots, real and imaginary, of y=-5(x-3)^2-45  using the quadratic formula?

Jul 4, 2017

$x = - \frac{3}{2} \pm \frac{9}{2} i$

#### Explanation:

To make this more strait forward lets convert the equation back to the form of $y = a {x}^{2} + b x + c$

Square the brackets.

$y = - 5 \left({x}^{2} - 3 x + 9\right) - 45$

Multiply out the -5

$y = - 5 {x}^{2} + 15 x - 45 - 45$

$y = - 5 {x}^{2} + 15 x - 90$

Thus the quadratic formula states that

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

where $a = - 5 \text{; "b=+15"; } c = - 90$

$x = \frac{- 15 \pm \sqrt{{\left(- 15\right)}^{2} - 4 \left(- 5\right) \left(- 90\right)}}{2 \left(- 5\right)}$

$x = - \frac{3}{2} \pm \frac{\sqrt{- 2025}}{- 10}$

If you are ever not sure about factors do a quick sketch of a prime factor tree on the side of the page. $x = - \frac{3}{2} \pm \frac{\sqrt{{5}^{2} \times {9}^{2} \times \left(- 1\right)}}{- 10}$

$x = - \frac{3}{2} \pm \frac{{\cancel{5}}^{1} \times 9 \times i}{{\cancel{10}}^{2}}$

As there is a $\pm$ sign in front of $\frac{9 \times i}{2}$ you can ignore any negative within it.

$x = - \frac{3}{2} \pm \frac{9}{2} i$