H=-16t^2+480(20)+150

Solution for H=-16t^2+480(20)+150 equation:

=-16H^2+480(20)+150

We move all terms to the left:

-(-16H^2+480(20)+150)=0

We get rid of parentheses

16H^2-48020-150=0

We add all the numbers together, and all the variables

16H^2-48170=0

a = 16; b = 0; c = -48170;
Δ = b2-4ac
Δ = 02-4·16·(-48170)
Δ = 3082880
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:

$\sqrt{\Delta}=\sqrt{3082880}=\sqrt{64*48170}=\sqrt{64}*\sqrt{48170}=8\sqrt{48170}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{48170}}{2*16}=\frac{0-8\sqrt{48170}}{32}
=-\frac{8\sqrt{48170}}{32}
=-\frac{\sqrt{48170}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{48170}}{2*16}=\frac{0+8\sqrt{48170}}{32}
=\frac{8\sqrt{48170}}{32}
=\frac{\sqrt{48170}}{4}
$