Y=-5t2+5t+30

Solution for Y=-5t2+5t+30 equation:

=-5Y^2+5Y+30

We move all terms to the left:

-(-5Y^2+5Y+30)=0

We get rid of parentheses

5Y^2-5Y-30=0

a = 5; b = -5; c = -30;
Δ = b2-4ac
Δ = -52-4·5·(-30)
Δ = 625
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{625}=25$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-25}{2*5}=\frac{-20}{10}
=-2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+25}{2*5}=\frac{30}{10}
=3
$