n2-3=3

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Solution for n2-3=3 equation:



n2-3=3
We move all terms to the left:
n2-3-(3)=0
We add all the numbers together, and all the variables
n^2-6=0
a = 1; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·1·(-6)
Δ = 24
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:
$\sqrt{\Delta}=\sqrt{24}=\sqrt{4*6}=\sqrt{4}*\sqrt{6}=2\sqrt{6}$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{6}}{2*1}=\frac{0-2\sqrt{6}}{2} =-\frac{2\sqrt{6}}{2} =-\sqrt{6} $
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{6}}{2*1}=\frac{0+2\sqrt{6}}{2} =\frac{2\sqrt{6}}{2} =\sqrt{6} $

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