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

Solution for 2x(3x-8)=3(4-x)+6x equation:

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

2x(3x-8)-(3(-1x+4)+6x)=0

We multiply parentheses

6x^2-16x-(3(-1x+4)+6x)=0


We calculate terms in parentheses: -(3(-1x+4)+6x), so:

3(-1x+4)+6x

We add all the numbers together, and all the variables

6x+3(-1x+4)

We multiply parentheses

6x-3x+12

We add all the numbers together, and all the variables

3x+12

Back to the equation:

-(3x+12)


We get rid of parentheses

6x^2-16x-3x-12=0

We add all the numbers together, and all the variables

6x^2-19x-12=0

a = 6; b = -19; c = -12;
Δ = b2-4ac
Δ = -192-4·6·(-12)
Δ = 649
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{649}}{2*6}=\frac{19-\sqrt{649}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{649}}{2*6}=\frac{19+\sqrt{649}}{12}
$