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

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

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

We move all terms to the left:

(1)-(-5H^2+10H)=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 = 80
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:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-4\sqrt{5}}{2*5}=\frac{10-4\sqrt{5}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+4\sqrt{5}}{2*5}=\frac{10+4\sqrt{5}}{10}
$