H=64+-3t+-5t2

Solution for H=64+-3t+-5t2 equation:

=64+-3H+-5H^2

We move all terms to the left:

-(64+-3H+-5H^2)=0

We use the square of the difference formula

-(64-3H-5H^2)=0

We get rid of parentheses

5H^2+3H-64=0

a = 5; b = 3; c = -64;
Δ = b2-4ac
Δ = 32-4·5·(-64)
Δ = 1289
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{1289}}{2*5}=\frac{-3-\sqrt{1289}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+\sqrt{1289}}{2*5}=\frac{-3+\sqrt{1289}}{10}
$