Y=-5t^2+20t+105

Solution for Y=-5t^2+20t+105 equation:

=-5Y^2+20Y+105

We move all terms to the left:

-(-5Y^2+20Y+105)=0

We get rid of parentheses

5Y^2-20Y-105=0

a = 5; b = -20; c = -105;
Δ = b2-4ac
Δ = -202-4·5·(-105)
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{-(-20)-50}{2*5}=\frac{-30}{10}
=-3
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+50}{2*5}=\frac{70}{10}
=7
$