H=3+90t-16t^2

Solution for H=3+90t-16t^2 equation:

=3+90H-16H^2

We move all terms to the left:

-(3+90H-16H^2)=0

We get rid of parentheses

16H^2-90H-3=0

a = 16; b = -90; c = -3;
Δ = b2-4ac
Δ = -902-4·16·(-3)
Δ = 8292
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:

$\sqrt{\Delta}=\sqrt{8292}=\sqrt{4*2073}=\sqrt{4}*\sqrt{2073}=2\sqrt{2073}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-90)-2\sqrt{2073}}{2*16}=\frac{90-2\sqrt{2073}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-90)+2\sqrt{2073}}{2*16}=\frac{90+2\sqrt{2073}}{32}
$