H=16t^2+80t+25

Solution for H=16t^2+80t+25 equation:

=16H^2+80H+25

We move all terms to the left:

-(16H^2+80H+25)=0

We get rid of parentheses

-16H^2-80H-25=0

a = -16; b = -80; c = -25;
Δ = b2-4ac
Δ = -802-4·(-16)·(-25)
Δ = 4800
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:

$\sqrt{\Delta}=\sqrt{4800}=\sqrt{1600*3}=\sqrt{1600}*\sqrt{3}=40\sqrt{3}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-80)-40\sqrt{3}}{2*-16}=\frac{80-40\sqrt{3}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-80)+40\sqrt{3}}{2*-16}=\frac{80+40\sqrt{3}}{-32}
$