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2x^2-512=0
a = 2; b = 0; c = -512;
Δ = b2-4ac
Δ = 02-4·2·(-512)
Δ = 4096
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{4096}=64$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-64}{2*2}=\frac{-64}{4} =-16 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+64}{2*2}=\frac{64}{4} =16 $
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