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2=2x^2
We move all terms to the left:
2-(2x^2)=0
a = -2; b = 0; c = +2;
Δ = b2-4ac
Δ = 02-4·(-2)·2
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4}{2*-2}=\frac{-4}{-4} =1 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4}{2*-2}=\frac{4}{-4} =-1 $
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