H=-5t^2+15+20

Solution for H=-5t^2+15+20 equation:

=-5H^2+15+20

We move all terms to the left:

-(-5H^2+15+20)=0

We get rid of parentheses

5H^2-15-20=0

We add all the numbers together, and all the variables

5H^2-35=0

a = 5; b = 0; c = -35;
Δ = b2-4ac
Δ = 02-4·5·(-35)
Δ = 700
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{700}=\sqrt{100*7}=\sqrt{100}*\sqrt{7}=10\sqrt{7}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{7}}{2*5}=\frac{0-10\sqrt{7}}{10}
=-\frac{10\sqrt{7}}{10}
=-\sqrt{7}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{7}}{2*5}=\frac{0+10\sqrt{7}}{10}
=\frac{10\sqrt{7}}{10}
=\sqrt{7}
$