Y=-5t^2+40t+45

Solution for Y=-5t^2+40t+45 equation:

=-5Y^2+40Y+45

We move all terms to the left:

-(-5Y^2+40Y+45)=0

We get rid of parentheses

5Y^2-40Y-45=0

a = 5; b = -40; c = -45;
Δ = b2-4ac
Δ = -402-4·5·(-45)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-50}{2*5}=\frac{-10}{10}
=-1
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+50}{2*5}=\frac{90}{10}
=9
$