H(t)=-5t^2+15t

Solution for H(t)=-5t^2+15t equation:

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

We move all terms to the left:

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

We get rid of parentheses

5H^2-15H+H=0

We add all the numbers together, and all the variables

5H^2-14H=0

a = 5; b = -14; c = 0;
Δ = b2-4ac
Δ = -142-4·5·0
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-14)-14}{2*5}=\frac{0}{10}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-14)+14}{2*5}=\frac{28}{10}
=2+4/5
$