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x^2-2+2x=0
a = 1; b = 2; c = -2;
Δ = b2-4ac
Δ = 22-4·1·(-2)
Δ = 12
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{12}=\sqrt{4*3}=\sqrt{4}*\sqrt{3}=2\sqrt{3}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{3}}{2*1}=\frac{-2-2\sqrt{3}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{3}}{2*1}=\frac{-2+2\sqrt{3}}{2} $
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