H=-5t2+10t+15

Solution for H=-5t2+10t+15 equation:

=-5H^2+10H+15

We move all terms to the left:

-(-5H^2+10H+15)=0

We get rid of parentheses

5H^2-10H-15=0

a = 5; b = -10; c = -15;
Δ = b2-4ac
Δ = -102-4·5·(-15)
Δ = 400
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{400}=20$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-20}{2*5}=\frac{-10}{10}
=-1
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+20}{2*5}=\frac{30}{10}
=3
$