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

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

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

We move all terms to the left:

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

We multiply parentheses

12x-(4(2x+3)4x)+8=0


We calculate terms in parentheses: -(4(2x+3)4x), so:

4(2x+3)4x

We multiply parentheses

32x^2+48x

Back to the equation:

-(32x^2+48x)


We get rid of parentheses

-32x^2+12x-48x+8=0

We add all the numbers together, and all the variables

-32x^2-36x+8=0

a = -32; b = -36; c = +8;
Δ = b2-4ac
Δ = -362-4·(-32)·8
Δ = 2320
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:

$\sqrt{\Delta}=\sqrt{2320}=\sqrt{16*145}=\sqrt{16}*\sqrt{145}=4\sqrt{145}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-36)-4\sqrt{145}}{2*-32}=\frac{36-4\sqrt{145}}{-64}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-36)+4\sqrt{145}}{2*-32}=\frac{36+4\sqrt{145}}{-64}
$