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

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

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

We move all terms to the left:

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

6x^2+5x-2=0

a = 6; b = 5; c = -2;
Δ = b2-4ac
Δ = 52-4·6·(-2)
Δ = 73
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{-(5)-\sqrt{73}}{2*6}=\frac{-5-\sqrt{73}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(5)+\sqrt{73}}{2*6}=\frac{-5+\sqrt{73}}{12}
$