2(3x+2)2x=28

Solution for 2(3x+2)2x=28 equation:

2(3x+2)2x=28

We move all terms to the left:

2(3x+2)2x-(28)=0

We multiply parentheses

12x^2+8x-28=0

a = 12; b = 8; c = -28;
Δ = b2-4ac
Δ = 82-4·12·(-28)
Δ = 1408
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:

$\sqrt{\Delta}=\sqrt{1408}=\sqrt{64*22}=\sqrt{64}*\sqrt{22}=8\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-8\sqrt{22}}{2*12}=\frac{-8-8\sqrt{22}}{24}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+8\sqrt{22}}{2*12}=\frac{-8+8\sqrt{22}}{24}
$