H=-5t^2+10t+1

Solution for H=-5t^2+10t+1 equation:

=-5H^2+10H+1

We move all terms to the left:

-(-5H^2+10H+1)=0

We get rid of parentheses

5H^2-10H-1=0

a = 5; b = -10; c = -1;
Δ = b2-4ac
Δ = -102-4·5·(-1)
Δ = 120
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:

$\sqrt{\Delta}=\sqrt{120}=\sqrt{4*30}=\sqrt{4}*\sqrt{30}=2\sqrt{30}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{30}}{2*5}=\frac{10-2\sqrt{30}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{30}}{2*5}=\frac{10+2\sqrt{30}}{10}
$