H=5t^2+20t+0.5

Solution for H=5t^2+20t+0.5 equation:

=5H^2+20H+0.5

We move all terms to the left:

-(5H^2+20H+0.5)=0

We get rid of parentheses

-5H^2-20H-0.5=0

a = -5; b = -20; c = -0.5;
Δ = b2-4ac
Δ = -202-4·(-5)·(-0.5)
Δ = 390
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{-(-20)-\sqrt{390}}{2*-5}=\frac{20-\sqrt{390}}{-10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+\sqrt{390}}{2*-5}=\frac{20+\sqrt{390}}{-10}
$