H=16t^2+84t+100

Solution for H=16t^2+84t+100 equation:

=16H^2+84H+100

We move all terms to the left:

-(16H^2+84H+100)=0

We get rid of parentheses

-16H^2-84H-100=0

a = -16; b = -84; c = -100;
Δ = b2-4ac
Δ = -842-4·(-16)·(-100)
Δ = 656
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:

$\sqrt{\Delta}=\sqrt{656}=\sqrt{16*41}=\sqrt{16}*\sqrt{41}=4\sqrt{41}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-84)-4\sqrt{41}}{2*-16}=\frac{84-4\sqrt{41}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-84)+4\sqrt{41}}{2*-16}=\frac{84+4\sqrt{41}}{-32}
$