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First I'll rewrite the equation properly:

2x²+3x=(3x-1)²-x²

Now we add x² to both sides and multiply out the (3x-1)²:

3x²+3x=9x²-6x+1

Then we subtract 3x² and 3x from both sides:

0=6x²-9x+1

Now using the quadratic equation (ax²+bx+c=0, y=(-b±√(b²-4ac))/2a):

(9±3√5)/12

So now we have:

(x-((9+3√5)/12))(x-((9-3√5)/12))=0

The zero product rule states ab=0 if a=0, b=0, or both of them =0, so if x= either (9+3√5)/12 or (9-3√5)/12 then the whole equation will equal 0.

So your answer is (9±3√5)/12

2x²+3x=(3x-1)²-x²

Now we add x² to both sides and multiply out the (3x-1)²:

3x²+3x=9x²-6x+1

Then we subtract 3x² and 3x from both sides:

0=6x²-9x+1

Now using the quadratic equation (ax²+bx+c=0, y=(-b±√(b²-4ac))/2a):

(9±3√5)/12

So now we have:

(x-((9+3√5)/12))(x-((9-3√5)/12))=0

The zero product rule states ab=0 if a=0, b=0, or both of them =0, so if x= either (9+3√5)/12 or (9-3√5)/12 then the whole equation will equal 0.

So your answer is (9±3√5)/12

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