(3x+1)(4x+2)=3

Solution for (3x+1)(4x+2)=3 equation:

(3x+1)(4x+2)=3

We move all terms to the left:

(3x+1)(4x+2)-(3)=0

We multiply parentheses ..

(+12x^2+6x+4x+2)-3=0

We get rid of parentheses

12x^2+6x+4x+2-3=0

We add all the numbers together, and all the variables

12x^2+10x-1=0

a = 12; b = 10; c = -1;
Δ = b2-4ac
Δ = 102-4·12·(-1)
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{-(10)-2\sqrt{37}}{2*12}=\frac{-10-2\sqrt{37}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(10)+2\sqrt{37}}{2*12}=\frac{-10+2\sqrt{37}}{24}
$