H=16t^2+112t+5

Solution for H=16t^2+112t+5 equation:

=16H^2+112H+5

We move all terms to the left:

-(16H^2+112H+5)=0

We get rid of parentheses

-16H^2-112H-5=0

a = -16; b = -112; c = -5;
Δ = b2-4ac
Δ = -1122-4·(-16)·(-5)
Δ = 12224
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:

$\sqrt{\Delta}=\sqrt{12224}=\sqrt{64*191}=\sqrt{64}*\sqrt{191}=8\sqrt{191}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-8\sqrt{191}}{2*-16}=\frac{112-8\sqrt{191}}{-32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+8\sqrt{191}}{2*-16}=\frac{112+8\sqrt{191}}{-32}
$