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6(n+2)n=2
We move all terms to the left:
6(n+2)n-(2)=0
We multiply parentheses
6n^2+12n-2=0
a = 6; b = 12; c = -2;
Δ = b2-4ac
Δ = 122-4·6·(-2)
Δ = 192
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-8\sqrt{3}}{2*6}=\frac{-12-8\sqrt{3}}{12} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+8\sqrt{3}}{2*6}=\frac{-12+8\sqrt{3}}{12} $
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