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(x+2)(x-2)-5(x+2)=0
We use the square of the difference formula
x^2-5(x+2)-4=0
We multiply parentheses
x^2-5x-10-4=0
We add all the numbers together, and all the variables
x^2-5x-14=0
a = 1; b = -5; c = -14;
Δ = b2-4ac
Δ = -52-4·1·(-14)
Δ = 81
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{81}=9$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-9}{2*1}=\frac{-4}{2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+9}{2*1}=\frac{14}{2} =7 $
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