H=-t^2+4t+50

Solution for H=-t^2+4t+50 equation:

=-H^2+4H+50

We move all terms to the left:

-(-H^2+4H+50)=0

We get rid of parentheses

H^2-4H-50=0

a = 1; b = -4; c = -50;
Δ = b2-4ac
Δ = -42-4·1·(-50)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-6\sqrt{6}}{2*1}=\frac{4-6\sqrt{6}}{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+6\sqrt{6}}{2*1}=\frac{4+6\sqrt{6}}{2}
$