H=38+-16t^2

Solution for H=38+-16t^2 equation:

=38+-16H^2

We move all terms to the left:

-(38+-16H^2)=0

We use the square of the difference formula

-(38-16H^2)=0

We get rid of parentheses

16H^2-38=0

a = 16; b = 0; c = -38;
Δ = b2-4ac
Δ = 02-4·16·(-38)
Δ = 2432
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:

$\sqrt{\Delta}=\sqrt{2432}=\sqrt{64*38}=\sqrt{64}*\sqrt{38}=8\sqrt{38}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{38}}{2*16}=\frac{0-8\sqrt{38}}{32}
=-\frac{8\sqrt{38}}{32}
=-\frac{\sqrt{38}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{38}}{2*16}=\frac{0+8\sqrt{38}}{32}
=\frac{8\sqrt{38}}{32}
=\frac{\sqrt{38}}{4}
$