# How do you find the roots, real and imaginary, of y=-(x -4 )^2-2x-1 using the quadratic formula?

Dec 27, 2017

${x}_{1} = 3 + 2 \sqrt{2} i$
${x}_{2} = 3 - 2 \sqrt{2} i$

#### Explanation:

$y = - {\left(x - 4\right)}^{2} - 2 x - 1 =$
$= - \left({x}^{2} - 8 x + 16\right) - 2 x - 1 =$
$= - {x}^{2} + 8 x - 16 - 2 x - 1 =$
$= - {x}^{2} + 6 x - 17$

Let $y = 0$
$\implies$
$- {x}^{2} + 6 x - 17 = 0 \iff a = - 1 , b = 6 , c = - 17$
$\implies$
${x}_{1 , 2} = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a} =$
$= \frac{- 6 \pm \sqrt{{6}^{2} - 4 \cdot \left(- 1\right) \cdot \left(- 17\right)}}{2 \left(- 1\right)} =$
$= \frac{- 6 \pm \sqrt{36 - 68}}{- 2} =$
$= \frac{- 6 \pm \sqrt{- 32}}{- 2} =$
$= \frac{- 6 \pm \sqrt{- \left(16 \cdot 2\right)}}{- 2} =$
$= \frac{- 6 \pm 4 \sqrt{2} i}{- 2} =$
$= \frac{\cancel{- 2} \left(3 \pm 2 \sqrt{2} i\right)}{\cancel{- 2}} =$
$= 3 \pm 2 \sqrt{2} i$
$\implies$
${x}_{1} = 3 + 2 \sqrt{2} i$
${x}_{2} = 3 - 2 \sqrt{2} i$