H=160+-96t+-16t2

Solution for H=160+-96t+-16t2 equation:

=160+-96H+-16H^2

We move all terms to the left:

-(160+-96H+-16H^2)=0

We use the square of the difference formula

-(160-96H-16H^2)=0

We get rid of parentheses

16H^2+96H-160=0

a = 16; b = 96; c = -160;
Δ = b2-4ac
Δ = 962-4·16·(-160)
Δ = 19456
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:

$\sqrt{\Delta}=\sqrt{19456}=\sqrt{1024*19}=\sqrt{1024}*\sqrt{19}=32\sqrt{19}$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(96)-32\sqrt{19}}{2*16}=\frac{-96-32\sqrt{19}}{32}
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(96)+32\sqrt{19}}{2*16}=\frac{-96+32\sqrt{19}}{32}
$