H=-16t^2+4t+132

Solution for H=-16t^2+4t+132 equation:

=-16H^2+4H+132

We move all terms to the left:

-(-16H^2+4H+132)=0

We get rid of parentheses

16H^2-4H-132=0

a = 16; b = -4; c = -132;
Δ = b2-4ac
Δ = -42-4·16·(-132)
Δ = 8464
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{8464}=92$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-92}{2*16}=\frac{-88}{32}
=-2+3/4
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+92}{2*16}=\frac{96}{32}
=3
$