192-112x+12x^2=0

Solution for 192-112x+12x^2=0 equation:

192-112x+12x^2=0

a = 12; b = -112; c = +192;
Δ = b2-4ac
Δ = -1122-4·12·192
Δ = 3328
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:

$\sqrt{\Delta}=\sqrt{3328}=\sqrt{256*13}=\sqrt{256}*\sqrt{13}=16\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-112)-16\sqrt{13}}{2*12}=\frac{112-16\sqrt{13}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-112)+16\sqrt{13}}{2*12}=\frac{112+16\sqrt{13}}{24}
$