H=-16t^2+130t+4

Solution for H=-16t^2+130t+4 equation:

=-16H^2+130H+4

We move all terms to the left:

-(-16H^2+130H+4)=0

We get rid of parentheses

16H^2-130H-4=0

a = 16; b = -130; c = -4;
Δ = b2-4ac
Δ = -1302-4·16·(-4)
Δ = 17156
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:

$\sqrt{\Delta}=\sqrt{17156}=\sqrt{4*4289}=\sqrt{4}*\sqrt{4289}=2\sqrt{4289}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-130)-2\sqrt{4289}}{2*16}=\frac{130-2\sqrt{4289}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-130)+2\sqrt{4289}}{2*16}=\frac{130+2\sqrt{4289}}{32}
$