H=16t^2+65t+10

Solution for H=16t^2+65t+10 equation:

=16H^2+65H+10

We move all terms to the left:

-(16H^2+65H+10)=0

We get rid of parentheses

-16H^2-65H-10=0

a = -16; b = -65; c = -10;
Δ = b2-4ac
Δ = -652-4·(-16)·(-10)
Δ = 3585
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{3585}}{2*-16}=\frac{65-\sqrt{3585}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-65)+\sqrt{3585}}{2*-16}=\frac{65+\sqrt{3585}}{-32}
$