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(3x+1)(3x-1)=35
We move all terms to the left:
(3x+1)(3x-1)-(35)=0
We use the square of the difference formula
9x^2-1-35=0
We add all the numbers together, and all the variables
9x^2-36=0
a = 9; b = 0; c = -36;
Δ = b2-4ac
Δ = 02-4·9·(-36)
Δ = 1296
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{1296}=36$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-36}{2*9}=\frac{-36}{18} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+36}{2*9}=\frac{36}{18} =2 $
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