H=-9t^2-15t+100

Solution for H=-9t^2-15t+100 equation:

=-9H^2-15H+100

We move all terms to the left:

-(-9H^2-15H+100)=0

We get rid of parentheses

9H^2+15H-100=0

a = 9; b = 15; c = -100;
Δ = b2-4ac
Δ = 152-4·9·(-100)
Δ = 3825
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:

$\sqrt{\Delta}=\sqrt{3825}=\sqrt{225*17}=\sqrt{225}*\sqrt{17}=15\sqrt{17}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-15\sqrt{17}}{2*9}=\frac{-15-15\sqrt{17}}{18}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+15\sqrt{17}}{2*9}=\frac{-15+15\sqrt{17}}{18}
$