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(2x-4)(2x+4)=0
We use the square of the difference formula
4x^2-16=0
a = 4; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·4·(-16)
Δ = 256
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{256}=16$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-16}{2*4}=\frac{-16}{8} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+16}{2*4}=\frac{16}{8} =2 $
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