H=-16t^2+52t+30

Solution for H=-16t^2+52t+30 equation:

=-16H^2+52H+30

We move all terms to the left:

-(-16H^2+52H+30)=0

We get rid of parentheses

16H^2-52H-30=0

a = 16; b = -52; c = -30;
Δ = b2-4ac
Δ = -522-4·16·(-30)
Δ = 4624
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4624}=68$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-52)-68}{2*16}=\frac{-16}{32}
=-1/2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-52)+68}{2*16}=\frac{120}{32}
=3+3/4
$