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5(x+1)(x-1)=2(3x-3)2-3x-1
We move all terms to the left:
5(x+1)(x-1)-(2(3x-3)2-3x-1)=0
We use the square of the difference formula
x^2-(2(3x-3)2-3x-1)-1=0
We calculate terms in parentheses: -(2(3x-3)2-3x-1), so:We get rid of parentheses
2(3x-3)2-3x-1
We add all the numbers together, and all the variables
-3x+2(3x-3)2-1
We multiply parentheses
-3x+12x-12-1
We add all the numbers together, and all the variables
9x-13
Back to the equation:
-(9x-13)
x^2-9x+13-1=0
We add all the numbers together, and all the variables
x^2-9x+12=0
a = 1; b = -9; c = +12;
Δ = b2-4ac
Δ = -92-4·1·12
Δ = 33
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-9)-\sqrt{33}}{2*1}=\frac{9-\sqrt{33}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-9)+\sqrt{33}}{2*1}=\frac{9+\sqrt{33}}{2} $
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