H(t)=-5t^2+20t+60

Solution for H(t)=-5t^2+20t+60 equation:

(H)=-5H^2+20H+60

We move all terms to the left:

(H)-(-5H^2+20H+60)=0

We get rid of parentheses

5H^2-20H+H-60=0

We add all the numbers together, and all the variables

5H^2-19H-60=0

a = 5; b = -19; c = -60;
Δ = b2-4ac
Δ = -192-4·5·(-60)
Δ = 1561
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{1561}}{2*5}=\frac{19-\sqrt{1561}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{1561}}{2*5}=\frac{19+\sqrt{1561}}{10}
$