H=-5t^2+20t+105

Solution for H=-5t^2+20t+105 equation:

=-5H^2+20H+105

We move all terms to the left:

-(-5H^2+20H+105)=0

We get rid of parentheses

5H^2-20H-105=0

a = 5; b = -20; c = -105;
Δ = b2-4ac
Δ = -202-4·5·(-105)
Δ = 2500
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2500}=50$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-20)-50}{2*5}=\frac{-30}{10}
=-3
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-20)+50}{2*5}=\frac{70}{10}
=7
$