H=-5t2+10t+35

Solution for H=-5t2+10t+35 equation:

=-5H^2+10H+35

We move all terms to the left:

-(-5H^2+10H+35)=0

We get rid of parentheses

5H^2-10H-35=0

a = 5; b = -10; c = -35;
Δ = b2-4ac
Δ = -102-4·5·(-35)
Δ = 800
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:

$\sqrt{\Delta}=\sqrt{800}=\sqrt{400*2}=\sqrt{400}*\sqrt{2}=20\sqrt{2}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-20\sqrt{2}}{2*5}=\frac{10-20\sqrt{2}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+20\sqrt{2}}{2*5}=\frac{10+20\sqrt{2}}{10}
$