H=196+-16t2

Solution for H=196+-16t2 equation:

=196+-16H^2

We move all terms to the left:

-(196+-16H^2)=0

We use the square of the difference formula

-(196-16H^2)=0

We get rid of parentheses

16H^2-196=0

a = 16; b = 0; c = -196;
Δ = b2-4ac
Δ = 02-4·16·(-196)
Δ = 12544
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{12544}=112$
$H_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-112}{2*16}=\frac{-112}{32}
=-3+1/2
$
$H_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+112}{2*16}=\frac{112}{32}
=3+1/2
$