H=48+8t-t^2

Solution for H=48+8t-t^2 equation:

=48+8H-H^2

We move all terms to the left:

-(48+8H-H^2)=0

We get rid of parentheses

H^2-8H-48=0

a = 1; b = -8; c = -48;
Δ = b2-4ac
Δ = -82-4·1·(-48)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*1}=\frac{-8}{2}
=-4
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*1}=\frac{24}{2}
=12
$