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

Solution for (2x-3)(3x-1)=(x+1)(2x+6) equation:

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

We move all terms to the left:

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

We multiply parentheses ..

(+6x^2-2x-9x+3)-((x+1)(2x+6))=0


We calculate terms in parentheses: -((x+1)(2x+6)), so:

(x+1)(2x+6)

We multiply parentheses ..

(+2x^2+6x+2x+6)

We get rid of parentheses

2x^2+6x+2x+6

We add all the numbers together, and all the variables

2x^2+8x+6

Back to the equation:

-(2x^2+8x+6)


We get rid of parentheses

6x^2-2x^2-2x-9x-8x+3-6=0

We add all the numbers together, and all the variables

4x^2-19x-3=0

a = 4; b = -19; c = -3;
Δ = b2-4ac
Δ = -192-4·4·(-3)
Δ = 409
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-19)-\sqrt{409}}{2*4}=\frac{19-\sqrt{409}}{8}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-19)+\sqrt{409}}{2*4}=\frac{19+\sqrt{409}}{8}
$