4(3x+1)32x=27

Solution for 4(3x+1)32x=27 equation:

4(3x+1)32x=27

We move all terms to the left:

4(3x+1)32x-(27)=0

We multiply parentheses

384x^2+128x-27=0

a = 384; b = 128; c = -27;
Δ = b2-4ac
Δ = 1282-4·384·(-27)
Δ = 57856
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:

$\sqrt{\Delta}=\sqrt{57856}=\sqrt{256*226}=\sqrt{256}*\sqrt{226}=16\sqrt{226}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(128)-16\sqrt{226}}{2*384}=\frac{-128-16\sqrt{226}}{768}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(128)+16\sqrt{226}}{2*384}=\frac{-128+16\sqrt{226}}{768}
$