H=3+38t-16t2

Solution for H=3+38t-16t2 equation:

=3+38H-16H^2

We move all terms to the left:

-(3+38H-16H^2)=0

We get rid of parentheses

16H^2-38H-3=0

a = 16; b = -38; c = -3;
Δ = b2-4ac
Δ = -382-4·16·(-3)
Δ = 1636
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:

$\sqrt{\Delta}=\sqrt{1636}=\sqrt{4*409}=\sqrt{4}*\sqrt{409}=2\sqrt{409}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-38)-2\sqrt{409}}{2*16}=\frac{38-2\sqrt{409}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-38)+2\sqrt{409}}{2*16}=\frac{38+2\sqrt{409}}{32}
$