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x(2x-4)=(x-2)(x-2)
We move all terms to the left:
x(2x-4)-((x-2)(x-2))=0
We multiply parentheses
2x^2-4x-((x-2)(x-2))=0
We multiply parentheses ..
2x^2-((+x^2-2x-2x+4))-4x=0
We calculate terms in parentheses: -((+x^2-2x-2x+4)), so:We add all the numbers together, and all the variables
(+x^2-2x-2x+4)
We get rid of parentheses
x^2-2x-2x+4
We add all the numbers together, and all the variables
x^2-4x+4
Back to the equation:
-(x^2-4x+4)
2x^2-4x-(x^2-4x+4)=0
We get rid of parentheses
2x^2-x^2-4x+4x-4=0
We add all the numbers together, and all the variables
x^2-4=0
a = 1; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·1·(-4)
Δ = 16
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{16}=4$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*1}=\frac{-4}{2} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*1}=\frac{4}{2} =2 $
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