H=2+65t-16t2

Solution for H=2+65t-16t2 equation:

=2+65H-16H^2

We move all terms to the left:

-(2+65H-16H^2)=0

We get rid of parentheses

16H^2-65H-2=0

a = 16; b = -65; c = -2;
Δ = b2-4ac
Δ = -652-4·16·(-2)
Δ = 4353
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{-(-65)-\sqrt{4353}}{2*16}=\frac{65-\sqrt{4353}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+\sqrt{4353}}{2*16}=\frac{65+\sqrt{4353}}{32}
$