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(2x-1)(2x-1)(x-1)(x+1)=0
We multiply parentheses ..
(+4x^2-2x-2x+1)(x-1)(x+1)=0
We use the square of the difference formula
x^2-1=0
a = 1; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·1·(-1)
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2}{2*1}=\frac{-2}{2} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2}{2*1}=\frac{2}{2} =1 $
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