H=-5+t^2+40t+45

Solution for H=-5+t^2+40t+45 equation:

=-5+H^2+40H+45

We move all terms to the left:

-(-5+H^2+40H+45)=0

We get rid of parentheses

-H^2-40H+5-45=0

We add all the numbers together, and all the variables

-1H^2-40H-40=0

a = -1; b = -40; c = -40;
Δ = b2-4ac
Δ = -402-4·(-1)·(-40)
Δ = 1440
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:

$\sqrt{\Delta}=\sqrt{1440}=\sqrt{144*10}=\sqrt{144}*\sqrt{10}=12\sqrt{10}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-40)-12\sqrt{10}}{2*-1}=\frac{40-12\sqrt{10}}{-2}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-40)+12\sqrt{10}}{2*-1}=\frac{40+12\sqrt{10}}{-2}
$