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

Solution for (2x-1)(3x+1)-(x+3)=0 equation:

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

We get rid of parentheses

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

We multiply parentheses ..

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

We add all the numbers together, and all the variables

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2-2x-4=0

a = 6; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·6·(-4)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-10}{2*6}=\frac{-8}{12}
=-2/3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+10}{2*6}=\frac{12}{12}
=1
$