(3x-12)x=96

Solution for (3x-12)x=96 equation:

(3x-12)x=96

We move all terms to the left:

(3x-12)x-(96)=0

We multiply parentheses

3x^2-12x-96=0

a = 3; b = -12; c = -96;
Δ = b2-4ac
Δ = -122-4·3·(-96)
Δ = 1296
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1296}=36$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-36}{2*3}=\frac{-24}{6}
=-4
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+36}{2*3}=\frac{48}{6}
=8
$