H=50+40t-5t^2

Solution for H=50+40t-5t^2 equation:

=50+40H-5H^2

We move all terms to the left:

-(50+40H-5H^2)=0

We get rid of parentheses

5H^2-40H-50=0

a = 5; b = -40; c = -50;
Δ = b2-4ac
Δ = -402-4·5·(-50)
Δ = 2600
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:

$\sqrt{\Delta}=\sqrt{2600}=\sqrt{100*26}=\sqrt{100}*\sqrt{26}=10\sqrt{26}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-10\sqrt{26}}{2*5}=\frac{40-10\sqrt{26}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+10\sqrt{26}}{2*5}=\frac{40+10\sqrt{26}}{10}
$