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

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

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

We move all terms to the left:

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

We multiply parentheses

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


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

2(4x+3)

We multiply parentheses

8x+6

Back to the equation:

-(8x+6)


We get rid of parentheses

6x^2+6x-8x-6=0

We add all the numbers together, and all the variables

6x^2-2x-6=0

a = 6; b = -2; c = -6;
Δ = b2-4ac
Δ = -22-4·6·(-6)
Δ = 148
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:

$\sqrt{\Delta}=\sqrt{148}=\sqrt{4*37}=\sqrt{4}*\sqrt{37}=2\sqrt{37}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{37}}{2*6}=\frac{2-2\sqrt{37}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{37}}{2*6}=\frac{2+2\sqrt{37}}{12}
$