2x(3x-4)=4x+3

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

2x(3x-4)=4x+3

We move all terms to the left:

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

We multiply parentheses

6x^2-8x-(4x+3)=0

We get rid of parentheses

6x^2-8x-4x-3=0

We add all the numbers together, and all the variables

6x^2-12x-3=0

a = 6; b = -12; c = -3;
Δ = b2-4ac
Δ = -122-4·6·(-3)
Δ = 216
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:

$\sqrt{\Delta}=\sqrt{216}=\sqrt{36*6}=\sqrt{36}*\sqrt{6}=6\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-6\sqrt{6}}{2*6}=\frac{12-6\sqrt{6}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+6\sqrt{6}}{2*6}=\frac{12+6\sqrt{6}}{12}
$