H=-16t^2+18t+15

Solution for H=-16t^2+18t+15 equation:

=-16H^2+18H+15

We move all terms to the left:

-(-16H^2+18H+15)=0

We get rid of parentheses

16H^2-18H-15=0

a = 16; b = -18; c = -15;
Δ = b2-4ac
Δ = -182-4·16·(-15)
Δ = 1284
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:

$\sqrt{\Delta}=\sqrt{1284}=\sqrt{4*321}=\sqrt{4}*\sqrt{321}=2\sqrt{321}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-2\sqrt{321}}{2*16}=\frac{18-2\sqrt{321}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+2\sqrt{321}}{2*16}=\frac{18+2\sqrt{321}}{32}
$