(3x)2+1=(3x+1)2

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

(3x)2+1=(3x+1)2

We move all terms to the left:

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

We add all the numbers together, and all the variables

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


We calculate terms in parentheses: -((3x+1)2), so:

(3x+1)2

We multiply parentheses

6x+2

Back to the equation:

-(6x+2)


We get rid of parentheses

3x^2-6x-2+1=0

We add all the numbers together, and all the variables

3x^2-6x-1=0

a = 3; b = -6; c = -1;
Δ = b2-4ac
Δ = -62-4·3·(-1)
Δ = 48
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:

$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-6)-4\sqrt{3}}{2*3}=\frac{6-4\sqrt{3}}{6}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+4\sqrt{3}}{2*3}=\frac{6+4\sqrt{3}}{6}
$