H=-16t^2+3t+64

Solution for H=-16t^2+3t+64 equation:

=-16H^2+3H+64

We move all terms to the left:

-(-16H^2+3H+64)=0

We get rid of parentheses

16H^2-3H-64=0

a = 16; b = -3; c = -64;
Δ = b2-4ac
Δ = -32-4·16·(-64)
Δ = 4105
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-\sqrt{4105}}{2*16}=\frac{3-\sqrt{4105}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+\sqrt{4105}}{2*16}=\frac{3+\sqrt{4105}}{32}
$