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(2/3)n+5=11
We move all terms to the left:
(2/3)n+5-(11)=0
Domain of the equation: 3)n!=0We add all the numbers together, and all the variables
n!=0/1
n!=0
n∈R
(+2/3)n+5-11=0
We add all the numbers together, and all the variables
(+2/3)n-6=0
We multiply parentheses
2n^2-6=0
a = 2; b = 0; c = -6;
Δ = b2-4ac
Δ = 02-4·2·(-6)
Δ = 48
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}$$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{48}=\sqrt{16*3}=\sqrt{16}*\sqrt{3}=4\sqrt{3}$$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{3}}{2*2}=\frac{0-4\sqrt{3}}{4} =-\frac{4\sqrt{3}}{4} =-\sqrt{3} $$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{3}}{2*2}=\frac{0+4\sqrt{3}}{4} =\frac{4\sqrt{3}}{4} =\sqrt{3} $
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