H=2+40-16t2

Solution for H=2+40-16t2 equation:

=2+40-16H^2

We move all terms to the left:

-(2+40-16H^2)=0

We get rid of parentheses

16H^2-2-40=0

We add all the numbers together, and all the variables

16H^2-42=0

a = 16; b = 0; c = -42;
Δ = b2-4ac
Δ = 02-4·16·(-42)
Δ = 2688
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:

$\sqrt{\Delta}=\sqrt{2688}=\sqrt{64*42}=\sqrt{64}*\sqrt{42}=8\sqrt{42}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{42}}{2*16}=\frac{0-8\sqrt{42}}{32}
=-\frac{8\sqrt{42}}{32}
=-\frac{\sqrt{42}}{4}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{42}}{2*16}=\frac{0+8\sqrt{42}}{32}
=\frac{8\sqrt{42}}{32}
=\frac{\sqrt{42}}{4}
$