H=40t+1-5t^2

Solution for H=40t+1-5t^2 equation:

=40H+1-5H^2

We move all terms to the left:

-(40H+1-5H^2)=0

We get rid of parentheses

5H^2-40H-1=0

a = 5; b = -40; c = -1;
Δ = b2-4ac
Δ = -402-4·5·(-1)
Δ = 1620
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:

$\sqrt{\Delta}=\sqrt{1620}=\sqrt{324*5}=\sqrt{324}*\sqrt{5}=18\sqrt{5}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-18\sqrt{5}}{2*5}=\frac{40-18\sqrt{5}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+18\sqrt{5}}{2*5}=\frac{40+18\sqrt{5}}{10}
$