H=-5t^2+10

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

=-5H^2+10

We move all terms to the left:

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

We get rid of parentheses

5H^2-10=0

a = 5; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·5·(-10)
Δ = 200
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:

$\sqrt{\Delta}=\sqrt{200}=\sqrt{100*2}=\sqrt{100}*\sqrt{2}=10\sqrt{2}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{2}}{2*5}=\frac{0-10\sqrt{2}}{10}
=-\frac{10\sqrt{2}}{10}
=-\sqrt{2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{2}}{2*5}=\frac{0+10\sqrt{2}}{10}
=\frac{10\sqrt{2}}{10}
=\sqrt{2}
$