112=-16x^2+96x

Solution for 112=-16x^2+96x equation:

112=-16x^2+96x

We move all terms to the left:

112-(-16x^2+96x)=0

We get rid of parentheses

16x^2-96x+112=0

a = 16; b = -96; c = +112;
Δ = b2-4ac
Δ = -962-4·16·112
Δ = 2048
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-32\sqrt{2}}{2*16}=\frac{96-32\sqrt{2}}{32}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+32\sqrt{2}}{2*16}=\frac{96+32\sqrt{2}}{32}
$