H=-5t^2+20+60

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

=-5H^2+20+60

We move all terms to the left:

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

We get rid of parentheses

5H^2-20-60=0

We add all the numbers together, and all the variables

5H^2-80=0

a = 5; b = 0; c = -80;
Δ = b2-4ac
Δ = 02-4·5·(-80)
Δ = 1600
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1600}=40$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-40}{2*5}=\frac{-40}{10}
=-4
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+40}{2*5}=\frac{40}{10}
=4
$