2(3^2x-5)-4=11

Solution for 2(3^2x-5)-4=11 equation:

2(3^2x-5)-4=11

We move all terms to the left:

2(3^2x-5)-4-(11)=0

We add all the numbers together, and all the variables

2(3^2x-5)-15=0

We multiply parentheses

6x^2-10-15=0

We add all the numbers together, and all the variables

6x^2-25=0

a = 6; b = 0; c = -25;
Δ = b2-4ac
Δ = 02-4·6·(-25)
Δ = 600
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:

$\sqrt{\Delta}=\sqrt{600}=\sqrt{100*6}=\sqrt{100}*\sqrt{6}=10\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-10\sqrt{6}}{2*6}=\frac{0-10\sqrt{6}}{12}
=-\frac{10\sqrt{6}}{12}
=-\frac{5\sqrt{6}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+10\sqrt{6}}{2*6}=\frac{0+10\sqrt{6}}{12}
=\frac{10\sqrt{6}}{12}
=\frac{5\sqrt{6}}{6}
$