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n2=42
We move all terms to the left:
n2-(42)=0
We add all the numbers together, and all the variables
n^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} =-\sqrt{42} $$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} =\sqrt{42} $
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