H=50+25t-5t^2

Solution for H=50+25t-5t^2 equation:

=50+25H-5H^2

We move all terms to the left:

-(50+25H-5H^2)=0

We get rid of parentheses

5H^2-25H-50=0

a = 5; b = -25; c = -50;
Δ = b2-4ac
Δ = -252-4·5·(-50)
Δ = 1625
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:

$\sqrt{\Delta}=\sqrt{1625}=\sqrt{25*65}=\sqrt{25}*\sqrt{65}=5\sqrt{65}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-25)-5\sqrt{65}}{2*5}=\frac{25-5\sqrt{65}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-25)+5\sqrt{65}}{2*5}=\frac{25+5\sqrt{65}}{10}
$