(x+1)=(2x+1)(3x)

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

(x+1)=(2x+1)(3x)

We move all terms to the left:

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

We get rid of parentheses

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


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

(2x+1)3x

We multiply parentheses

6x^2+3x

Back to the equation:

-(6x^2+3x)


We get rid of parentheses

-6x^2+x-3x+1=0

We add all the numbers together, and all the variables

-6x^2-2x+1=0

a = -6; b = -2; c = +1;
Δ = b2-4ac
Δ = -22-4·(-6)·1
Δ = 28
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:

$\sqrt{\Delta}=\sqrt{28}=\sqrt{4*7}=\sqrt{4}*\sqrt{7}=2\sqrt{7}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2\sqrt{7}}{2*-6}=\frac{2-2\sqrt{7}}{-12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2\sqrt{7}}{2*-6}=\frac{2+2\sqrt{7}}{-12}
$