3(192+-32x+x^2)=0

Solution for 3(192+-32x+x^2)=0 equation:

3(192+-32x+x^2)=0

We use the square of the difference formula

3(192-32x+x^2)=0

We multiply parentheses

3x^2-96x+576=0

a = 3; b = -96; c = +576;
Δ = b2-4ac
Δ = -962-4·3·576
Δ = 2304
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{2304}=48$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-96)-48}{2*3}=\frac{48}{6}
=8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-96)+48}{2*3}=\frac{144}{6}
=24
$