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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

4(2x+3)

We multiply parentheses

8x+12

Back to the equation:

-(8x+12)


We add all the numbers together, and all the variables

3x^2-10x-(8x+12)=0

We get rid of parentheses

3x^2-10x-8x-12=0

We add all the numbers together, and all the variables

3x^2-18x-12=0

a = 3; b = -18; c = -12;
Δ = b2-4ac
Δ = -182-4·3·(-12)
Δ = 468
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:

$\sqrt{\Delta}=\sqrt{468}=\sqrt{36*13}=\sqrt{36}*\sqrt{13}=6\sqrt{13}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-18)-6\sqrt{13}}{2*3}=\frac{18-6\sqrt{13}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-18)+6\sqrt{13}}{2*3}=\frac{18+6\sqrt{13}}{6}
$