H=-5t^2+40t+2

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

=-5H^2+40H+2

We move all terms to the left:

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

We get rid of parentheses

5H^2-40H-2=0

a = 5; b = -40; c = -2;
Δ = b2-4ac
Δ = -402-4·5·(-2)
Δ = 1640
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:

$\sqrt{\Delta}=\sqrt{1640}=\sqrt{4*410}=\sqrt{4}*\sqrt{410}=2\sqrt{410}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-2\sqrt{410}}{2*5}=\frac{40-2\sqrt{410}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+2\sqrt{410}}{2*5}=\frac{40+2\sqrt{410}}{10}
$