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6+n^2=16
We move all terms to the left:
6+n^2-(16)=0
We add all the numbers together, and all the variables
n^2-10=0
a = 1; b = 0; c = -10;
Δ = b2-4ac
Δ = 02-4·1·(-10)
Δ = 40
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{40}=\sqrt{4*10}=\sqrt{4}*\sqrt{10}=2\sqrt{10}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{10}}{2*1}=\frac{0-2\sqrt{10}}{2} =-\frac{2\sqrt{10}}{2} =-\sqrt{10} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{10}}{2*1}=\frac{0+2\sqrt{10}}{2} =\frac{2\sqrt{10}}{2} =\sqrt{10} $
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