H(t)=-5t^2+20t+4

Solution for H(t)=-5t^2+20t+4 equation:

(H)=-5H^2+20H+4

We move all terms to the left:

(H)-(-5H^2+20H+4)=0

We get rid of parentheses

5H^2-20H+H-4=0

We add all the numbers together, and all the variables

5H^2-19H-4=0

a = 5; b = -19; c = -4;
Δ = b2-4ac
Δ = -192-4·5·(-4)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-21}{2*5}=\frac{-2}{10}
=-1/5
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+21}{2*5}=\frac{40}{10}
=4
$