2(x+1)3x=17

Solution for 2(x+1)3x=17 equation:

2(x+1)3x=17

We move all terms to the left:

2(x+1)3x-(17)=0

We multiply parentheses

6x^2+6x-17=0

a = 6; b = 6; c = -17;
Δ = b2-4ac
Δ = 62-4·6·(-17)
Δ = 444
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:

$\sqrt{\Delta}=\sqrt{444}=\sqrt{4*111}=\sqrt{4}*\sqrt{111}=2\sqrt{111}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-2\sqrt{111}}{2*6}=\frac{-6-2\sqrt{111}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{111}}{2*6}=\frac{-6+2\sqrt{111}}{12}
$