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4=2x^2
We move all terms to the left:
4-(2x^2)=0
a = -2; b = 0; c = +4;
Δ = b2-4ac
Δ = 02-4·(-2)·4
Δ = 32
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{32}=\sqrt{16*2}=\sqrt{16}*\sqrt{2}=4\sqrt{2}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{2}}{2*-2}=\frac{0-4\sqrt{2}}{-4} =-\frac{4\sqrt{2}}{-4} =-\frac{\sqrt{2}}{-1} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{2}}{2*-2}=\frac{0+4\sqrt{2}}{-4} =\frac{4\sqrt{2}}{-4} =\frac{\sqrt{2}}{-1} $
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