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(2x+4)(2x-2)=16
We move all terms to the left:
(2x+4)(2x-2)-(16)=0
We multiply parentheses ..
(+4x^2-4x+8x-8)-16=0
We get rid of parentheses
4x^2-4x+8x-8-16=0
We add all the numbers together, and all the variables
4x^2+4x-24=0
a = 4; b = 4; c = -24;
Δ = b2-4ac
Δ = 42-4·4·(-24)
Δ = 400
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{400}=20$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-20}{2*4}=\frac{-24}{8} =-3 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+20}{2*4}=\frac{16}{8} =2 $
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