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2x^2=8
We move all terms to the left:
2x^2-(8)=0
a = 2; b = 0; c = -8;
Δ = b2-4ac
Δ = 02-4·2·(-8)
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*2}=\frac{-8}{4} =-2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*2}=\frac{8}{4} =2 $
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