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(2n-1)(2n+1)=0
We use the square of the difference formula
4n^2-1=0
a = 4; b = 0; c = -1;
Δ = b2-4ac
Δ = 02-4·4·(-1)
Δ = 16
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{16}=4$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*4}=\frac{-4}{8} =-1/2 $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*4}=\frac{4}{8} =1/2 $
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