H=90+15t+-5t2

Solution for H=90+15t+-5t2 equation:

=90+15H+-5H^2

We move all terms to the left:

-(90+15H+-5H^2)=0

We use the square of the difference formula

-(90+15H-5H^2)=0

We get rid of parentheses

5H^2-15H-90=0

a = 5; b = -15; c = -90;
Δ = b2-4ac
Δ = -152-4·5·(-90)
Δ = 2025
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2025}=45$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-45}{2*5}=\frac{-30}{10}
=-3
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+45}{2*5}=\frac{60}{10}
=6
$