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(x+4)(x+8)=2x(x+4)
We move all terms to the left:
(x+4)(x+8)-(2x(x+4))=0
We multiply parentheses ..
(+x^2+8x+4x+32)-(2x(x+4))=0
We calculate terms in parentheses: -(2x(x+4)), so:We get rid of parentheses
2x(x+4)
We multiply parentheses
2x^2+8x
Back to the equation:
-(2x^2+8x)
x^2-2x^2+8x+4x-8x+32=0
We add all the numbers together, and all the variables
-1x^2+4x+32=0
a = -1; b = 4; c = +32;
Δ = b2-4ac
Δ = 42-4·(-1)·32
Δ = 144
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{144}=12$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-12}{2*-1}=\frac{-16}{-2} =+8 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+12}{2*-1}=\frac{8}{-2} =-4 $
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