H(t)=112t+-16t^2

Solution for H(t)=112t+-16t^2 equation:

(H)=112H+-16H^2

We move all terms to the left:

(H)-(112H+-16H^2)=0

We use the square of the difference formula

-(112H-16H^2)+H=0

We get rid of parentheses

16H^2-112H+H=0

We add all the numbers together, and all the variables

16H^2-111H=0

a = 16; b = -111; c = 0;
Δ = b2-4ac
Δ = -1112-4·16·0
Δ = 12321
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12321}=111$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-111)-111}{2*16}=\frac{0}{32}
=0
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-111)+111}{2*16}=\frac{222}{32}
=6+15/16
$