200=112+96(t)-16(t)^2

Solution for 200=112+96(t)-16(t)^2 equation:

200=112+96(t)-16(t)^2

We move all terms to the left:

200-(112+96(t)-16(t)^2)=0

We get rid of parentheses

16t^2-96t-112+200=0

We add all the numbers together, and all the variables

16t^2-96t+88=0

a = 16; b = -96; c = +88;
Δ = b2-4ac
Δ = -962-4·16·88
Δ = 3584
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{3584}=\sqrt{256*14}=\sqrt{256}*\sqrt{14}=16\sqrt{14}$
$t_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-16\sqrt{14}}{2*16}=\frac{96-16\sqrt{14}}{32}
$
$t_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+16\sqrt{14}}{2*16}=\frac{96+16\sqrt{14}}{32}
$