3x^2-4x=2(x+1)

Solution for 3x^2-4x=2(x+1) equation:

3x^2-4x=2(x+1)

We move all terms to the left:

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


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

2(x+1)

We multiply parentheses

2x+2

Back to the equation:

-(2x+2)


We get rid of parentheses

3x^2-4x-2x-2=0

We add all the numbers together, and all the variables

3x^2-6x-2=0

a = 3; b = -6; c = -2;
Δ = b2-4ac
Δ = -62-4·3·(-2)
Δ = 60
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:

$\sqrt{\Delta}=\sqrt{60}=\sqrt{4*15}=\sqrt{4}*\sqrt{15}=2\sqrt{15}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-2\sqrt{15}}{2*3}=\frac{6-2\sqrt{15}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+2\sqrt{15}}{2*3}=\frac{6+2\sqrt{15}}{6}
$