H=-16t2+90t+20

Solution for H=-16t2+90t+20 equation:

=-16H^2+90H+20

We move all terms to the left:

-(-16H^2+90H+20)=0

We get rid of parentheses

16H^2-90H-20=0

a = 16; b = -90; c = -20;
Δ = b2-4ac
Δ = -902-4·16·(-20)
Δ = 9380
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:

$\sqrt{\Delta}=\sqrt{9380}=\sqrt{4*2345}=\sqrt{4}*\sqrt{2345}=2\sqrt{2345}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-2\sqrt{2345}}{2*16}=\frac{90-2\sqrt{2345}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+2\sqrt{2345}}{2*16}=\frac{90+2\sqrt{2345}}{32}
$