0=16x^2+96x+112

Solution for 0=16x^2+96x+112 equation:

0=16x^2+96x+112

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(16x^2+96x+112)=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}
$