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

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

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

We move all terms to the left:

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

We add all the numbers together, and all the variables

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

We multiply parentheses

12x^2-18x-(3(-1x+4))=0


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

3(-1x+4)

We multiply parentheses

-3x+12

Back to the equation:

-(-3x+12)


We get rid of parentheses

12x^2-18x+3x-12=0

We add all the numbers together, and all the variables

12x^2-15x-12=0

a = 12; b = -15; c = -12;
Δ = b2-4ac
Δ = -152-4·12·(-12)
Δ = 801
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:

$\sqrt{\Delta}=\sqrt{801}=\sqrt{9*89}=\sqrt{9}*\sqrt{89}=3\sqrt{89}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-15)-3\sqrt{89}}{2*12}=\frac{15-3\sqrt{89}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-15)+3\sqrt{89}}{2*12}=\frac{15+3\sqrt{89}}{24}
$