H=-10t^2+20t+105

Solution for H=-10t^2+20t+105 equation:

=-10H^2+20H+105

We move all terms to the left:

-(-10H^2+20H+105)=0

We get rid of parentheses

10H^2-20H-105=0

a = 10; b = -20; c = -105;
Δ = b2-4ac
Δ = -202-4·10·(-105)
Δ = 4600
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:

$\sqrt{\Delta}=\sqrt{4600}=\sqrt{100*46}=\sqrt{100}*\sqrt{46}=10\sqrt{46}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-10\sqrt{46}}{2*10}=\frac{20-10\sqrt{46}}{20}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+10\sqrt{46}}{2*10}=\frac{20+10\sqrt{46}}{20}
$