3x(2x-4)=16

Solution for 3x(2x-4)=16 equation:

3x(2x-4)=16

We move all terms to the left:

3x(2x-4)-(16)=0

We multiply parentheses

6x^2-12x-16=0

a = 6; b = -12; c = -16;
Δ = b2-4ac
Δ = -122-4·6·(-16)
Δ = 528
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:

$\sqrt{\Delta}=\sqrt{528}=\sqrt{16*33}=\sqrt{16}*\sqrt{33}=4\sqrt{33}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-4\sqrt{33}}{2*6}=\frac{12-4\sqrt{33}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+4\sqrt{33}}{2*6}=\frac{12+4\sqrt{33}}{12}
$