H(t)=-5t^2+40t+12

Solution for H(t)=-5t^2+40t+12 equation:

(H)=-5H^2+40H+12

We move all terms to the left:

(H)-(-5H^2+40H+12)=0

We get rid of parentheses

5H^2-40H+H-12=0

We add all the numbers together, and all the variables

5H^2-39H-12=0

a = 5; b = -39; c = -12;
Δ = b2-4ac
Δ = -392-4·5·(-12)
Δ = 1761
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-39)-\sqrt{1761}}{2*5}=\frac{39-\sqrt{1761}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-39)+\sqrt{1761}}{2*5}=\frac{39+\sqrt{1761}}{10}
$