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+3=6N^2
We move all terms to the left:
+3-(6N^2)=0
a = -6; b = 0; c = +3;
Δ = b2-4ac
Δ = 02-4·(-6)·3
Δ = 72
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{72}=\sqrt{36*2}=\sqrt{36}*\sqrt{2}=6\sqrt{2}$$N_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{2}}{2*-6}=\frac{0-6\sqrt{2}}{-12} =-\frac{6\sqrt{2}}{-12} =-\frac{\sqrt{2}}{-2} $$N_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{2}}{2*-6}=\frac{0+6\sqrt{2}}{-12} =\frac{6\sqrt{2}}{-12} =\frac{\sqrt{2}}{-2} $
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