H=-5t^2+100+15

Solution for H=-5t^2+100+15 equation:

=-5H^2+100+15

We move all terms to the left:

-(-5H^2+100+15)=0

We get rid of parentheses

5H^2-100-15=0

We add all the numbers together, and all the variables

5H^2-115=0

a = 5; b = 0; c = -115;
Δ = b2-4ac
Δ = 02-4·5·(-115)
Δ = 2300
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:

$\sqrt{\Delta}=\sqrt{2300}=\sqrt{100*23}=\sqrt{100}*\sqrt{23}=10\sqrt{23}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{23}}{2*5}=\frac{0-10\sqrt{23}}{10}
=-\frac{10\sqrt{23}}{10}
=-\sqrt{23}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{23}}{2*5}=\frac{0+10\sqrt{23}}{10}
=\frac{10\sqrt{23}}{10}
=\sqrt{23}
$