H=-5t^2+100t+350

Solution for H=-5t^2+100t+350 equation:

=-5H^2+100H+350

We move all terms to the left:

-(-5H^2+100H+350)=0

We get rid of parentheses

5H^2-100H-350=0

a = 5; b = -100; c = -350;
Δ = b2-4ac
Δ = -1002-4·5·(-350)
Δ = 17000
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:

$\sqrt{\Delta}=\sqrt{17000}=\sqrt{100*170}=\sqrt{100}*\sqrt{170}=10\sqrt{170}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-10\sqrt{170}}{2*5}=\frac{100-10\sqrt{170}}{10}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+10\sqrt{170}}{2*5}=\frac{100+10\sqrt{170}}{10}
$