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2x+(1/6)x=13
We move all terms to the left:
2x+(1/6)x-(13)=0
Domain of the equation: 6)x!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
2x+(+1/6)x-13=0
We multiply parentheses
x^2+2x-13=0
a = 1; b = 2; c = -13;
Δ = b2-4ac
Δ = 22-4·1·(-13)
Δ = 56
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{56}=\sqrt{4*14}=\sqrt{4}*\sqrt{14}=2\sqrt{14}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{14}}{2*1}=\frac{-2-2\sqrt{14}}{2} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{14}}{2*1}=\frac{-2+2\sqrt{14}}{2} $
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