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39=n2-3
We move all terms to the left:
39-(n2-3)=0
We add all the numbers together, and all the variables
-(+n^2-3)+39=0
We get rid of parentheses
-n^2+3+39=0
We add all the numbers together, and all the variables
-1n^2+42=0
a = -1; b = 0; c = +42;
Δ = b2-4ac
Δ = 02-4·(-1)·42
Δ = 168
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{168}=\sqrt{4*42}=\sqrt{4}*\sqrt{42}=2\sqrt{42}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{42}}{2*-1}=\frac{0-2\sqrt{42}}{-2} =-\frac{2\sqrt{42}}{-2} =-\frac{\sqrt{42}}{-1} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{42}}{2*-1}=\frac{0+2\sqrt{42}}{-2} =\frac{2\sqrt{42}}{-2} =\frac{\sqrt{42}}{-1} $
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