Y=-5t^2+15t+50

Solution for Y=-5t^2+15t+50 equation:

=-5Y^2+15Y+50

We move all terms to the left:

-(-5Y^2+15Y+50)=0

We get rid of parentheses

5Y^2-15Y-50=0

a = 5; b = -15; c = -50;
Δ = b2-4ac
Δ = -152-4·5·(-50)
Δ = 1225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1225}=35$
$Y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-35}{2*5}=\frac{-20}{10}
=-2
$
$Y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+35}{2*5}=\frac{50}{10}
=5
$