H(t)=-3t^2+3t+150

Solution for H(t)=-3t^2+3t+150 equation:

(H)=-3H^2+3H+150

We move all terms to the left:

(H)-(-3H^2+3H+150)=0

We get rid of parentheses

3H^2-3H+H-150=0

We add all the numbers together, and all the variables

3H^2-2H-150=0

a = 3; b = -2; c = -150;
Δ = b2-4ac
Δ = -22-4·3·(-150)
Δ = 1804
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:

$\sqrt{\Delta}=\sqrt{1804}=\sqrt{4*451}=\sqrt{4}*\sqrt{451}=2\sqrt{451}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{451}}{2*3}=\frac{2-2\sqrt{451}}{6}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{451}}{2*3}=\frac{2+2\sqrt{451}}{6}
$