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2x^2-4=2x
We move all terms to the left:
2x^2-4-(2x)=0
a = 2; b = -2; c = -4;
Δ = b2-4ac
Δ = -22-4·2·(-4)
Δ = 36
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{36}=6$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-6}{2*2}=\frac{-4}{4} =-1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+6}{2*2}=\frac{8}{4} =2 $
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