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8=2b^2
We move all terms to the left:
8-(2b^2)=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:$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{64}=8$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8}{2*-2}=\frac{-8}{-4} =+2 $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8}{2*-2}=\frac{8}{-4} =-2 $
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