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b^2=512
We move all terms to the left:
b^2-(512)=0
a = 1; b = 0; c = -512;
Δ = b2-4ac
Δ = 02-4·1·(-512)
Δ = 2048
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}$$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{2048}=\sqrt{1024*2}=\sqrt{1024}*\sqrt{2}=32\sqrt{2}$$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-32\sqrt{2}}{2*1}=\frac{0-32\sqrt{2}}{2} =-\frac{32\sqrt{2}}{2} =-16\sqrt{2} $$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+32\sqrt{2}}{2*1}=\frac{0+32\sqrt{2}}{2} =\frac{32\sqrt{2}}{2} =16\sqrt{2} $
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