# How do you find the roots, real and imaginary, of y=-(x -2)^2-4x+5 using the quadratic formula?

the answers are $y = \pm 1$
the given equation is $y = - {\left(x - 2\right)}^{2} - 4 x + 5 \Rightarrow y = - {x}^{2} - 4 + 4 x - 4 x + 5 \Rightarrow y = - {x}^{2} + 1$ on applying the quadratic formula $\textcolor{red}{y = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}}$ here $a = - 1 , b = 0 , c = 1$ on substituting the values we get $\frac{0 \pm \sqrt{0 - 4 \left(- 1\right) \left(1\right)}}{2 \left(1\right)} \Rightarrow \left(\pm \frac{\sqrt{4}}{2}\right) \Rightarrow \pm \frac{2}{2} \Rightarrow \pm 1$
$\therefore$ $y = \pm 1$