H=-5t^2+20t+0.6

Solution for H=-5t^2+20t+0.6 equation:

=-5H^2+20H+0.6

We move all terms to the left:

-(-5H^2+20H+0.6)=0

We get rid of parentheses

5H^2-20H-0.6=0

a = 5; b = -20; c = -0.6;
Δ = b2-4ac
Δ = -202-4·5·(-0.6)
Δ = 412
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:

$\sqrt{\Delta}=\sqrt{412}=\sqrt{4*103}=\sqrt{4}*\sqrt{103}=2\sqrt{103}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-2\sqrt{103}}{2*5}=\frac{20-2\sqrt{103}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+2\sqrt{103}}{2*5}=\frac{20+2\sqrt{103}}{10}
$