H(t)=50-5t-5t^2

Solution for H(t)=50-5t-5t^2 equation:

(H)=50-5H-5H^2

We move all terms to the left:

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

We get rid of parentheses

5H^2+5H+H-50=0

We add all the numbers together, and all the variables

5H^2+6H-50=0

a = 5; b = 6; c = -50;
Δ = b2-4ac
Δ = 62-4·5·(-50)
Δ = 1036
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:

$\sqrt{\Delta}=\sqrt{1036}=\sqrt{4*259}=\sqrt{4}*\sqrt{259}=2\sqrt{259}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{259}}{2*5}=\frac{-6-2\sqrt{259}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{259}}{2*5}=\frac{-6+2\sqrt{259}}{10}
$