We must find the inverse of:$ h(t) = -5t^2+30t+60 $ ...
$0= -5t^2+30t+60-h $
We use quadratic formula. I am trying to understand how the denominator plays in to the solution presented: $$t = \frac{−30 ± \sqrt{30^2+20(60 − h)}}{-10} $$ $$= {3 ± \sqrt{3^2+.2(60 − h)}} $$ ... and so on.
I see that we divide out the denominator from $ -b $ in the quadratic formula, I dont understand why $3^2$ remains positive having been divided by a negative ($-10$), and how we get $0.2$ from dividing $20$ in the numberator by $-10$.
Also, why would the same reduction not apply to the coefficient $60$ in $(60-h)$ ?
any insight is appreciated. thanks!
The impact of dividing by a negative number is shown below.
\begin{align}t &= \frac{−30 \pm \sqrt{30^2+20(60 − h)}}{-10}\\ &= {3 \color{red}{\mp} \sqrt{\frac{30^2+20(60 − h)}{10^2}}} \end{align}
For example $\frac{1+\sqrt{2}}{-1}=-1-\sqrt{2}$ and $\frac{1-\sqrt{2}}{-1}=-1+\sqrt{2}$