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(X-4)(4+X)=0
We add all the numbers together, and all the variables
(X-4)(X+4)=0
We use the square of the difference formula
X^2-16=0
a = 1; b = 0; c = -16;
Δ = b2-4ac
Δ = 02-4·1·(-16)
Δ = 64
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{64}=8$$X_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*1}=\frac{-8}{2} =-4 $$X_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*1}=\frac{8}{2} =4 $
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