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x^2-4=2(x^-2)
We move all terms to the left:
x^2-4-(2(x^-2))=0
We calculate terms in parentheses: -(2(x^-2)), so:We get rid of parentheses
2(x^-2)
We multiply parentheses
2x-4
Back to the equation:
-(2x-4)
x^2-2x+4-4=0
We add all the numbers together, and all the variables
x^2-2x=0
a = 1; b = -2; c = 0;
Δ = b2-4ac
Δ = -22-4·1·0
Δ = 4
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{4}=2$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-2}{2*1}=\frac{0}{2} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+2}{2*1}=\frac{4}{2} =2 $
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