3x(2x-4)+3x=3(2x-4)+2(5-x)

Solution for 3x(2x-4)+3x=3(2x-4)+2(5-x) equation:

3x(2x-4)+3x=3(2x-4)+2(5-x)

We move all terms to the left:

3x(2x-4)+3x-(3(2x-4)+2(5-x))=0

We add all the numbers together, and all the variables

3x(2x-4)+3x-(3(2x-4)+2(-1x+5))=0

We add all the numbers together, and all the variables

3x+3x(2x-4)-(3(2x-4)+2(-1x+5))=0

We multiply parentheses

6x^2+3x-12x-(3(2x-4)+2(-1x+5))=0


We calculate terms in parentheses: -(3(2x-4)+2(-1x+5)), so:

3(2x-4)+2(-1x+5)

We multiply parentheses

6x-2x-12+10

We add all the numbers together, and all the variables

4x-2

Back to the equation:

-(4x-2)


We add all the numbers together, and all the variables

6x^2-9x-(4x-2)=0

We get rid of parentheses

6x^2-9x-4x+2=0

We add all the numbers together, and all the variables

6x^2-13x+2=0

a = 6; b = -13; c = +2;
Δ = b2-4ac
Δ = -132-4·6·2
Δ = 121
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{121}=11$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-13)-11}{2*6}=\frac{2}{12}
=1/6
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-13)+11}{2*6}=\frac{24}{12}
=2
$