H(t)=-5t^2+20t+12

Solution for H(t)=-5t^2+20t+12 equation:

(H)=-5H^2+20H+12

We move all terms to the left:

(H)-(-5H^2+20H+12)=0

We get rid of parentheses

5H^2-20H+H-12=0

We add all the numbers together, and all the variables

5H^2-19H-12=0

a = 5; b = -19; c = -12;
Δ = b2-4ac
Δ = -192-4·5·(-12)
Δ = 601
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{-(-19)-\sqrt{601}}{2*5}=\frac{19-\sqrt{601}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{601}}{2*5}=\frac{19+\sqrt{601}}{10}
$