H(t)=-5t^2+150t+100

Solution for H(t)=-5t^2+150t+100 equation:

(H)=-5H^2+150H+100

We move all terms to the left:

(H)-(-5H^2+150H+100)=0

We get rid of parentheses

5H^2-150H+H-100=0

We add all the numbers together, and all the variables

5H^2-149H-100=0

a = 5; b = -149; c = -100;
Δ = b2-4ac
Δ = -1492-4·5·(-100)
Δ = 24201
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:

$\sqrt{\Delta}=\sqrt{24201}=\sqrt{9*2689}=\sqrt{9}*\sqrt{2689}=3\sqrt{2689}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-149)-3\sqrt{2689}}{2*5}=\frac{149-3\sqrt{2689}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-149)+3\sqrt{2689}}{2*5}=\frac{149+3\sqrt{2689}}{10}
$