(3x)^2=6+4x+22

Solution for (3x)^2=6+4x+22 equation:

(3x)^2=6+4x+22

We move all terms to the left:

(3x)^2-(6+4x+22)=0

We add all the numbers together, and all the variables

3x^2-(4x+28)=0

We get rid of parentheses

3x^2-4x-28=0

a = 3; b = -4; c = -28;
Δ = b2-4ac
Δ = -42-4·3·(-28)
Δ = 352
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:

$\sqrt{\Delta}=\sqrt{352}=\sqrt{16*22}=\sqrt{16}*\sqrt{22}=4\sqrt{22}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-4\sqrt{22}}{2*3}=\frac{4-4\sqrt{22}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+4\sqrt{22}}{2*3}=\frac{4+4\sqrt{22}}{6}
$