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(3x+1)2=9x^2+6x+1
We move all terms to the left:
(3x+1)2-(9x^2+6x+1)=0
We multiply parentheses
6x-(9x^2+6x+1)+2=0
We get rid of parentheses
-9x^2+6x-6x-1+2=0
We add all the numbers together, and all the variables
-9x^2+1=0
a = -9; b = 0; c = +1;
Δ = b2-4ac
Δ = 02-4·(-9)·1
Δ = 36
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6}{2*-9}=\frac{-6}{-18} =1/3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6}{2*-9}=\frac{6}{-18} =-1/3 $
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