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2z+6=1/3z+1
We move all terms to the left:
2z+6-(1/3z+1)=0
Domain of the equation: 3z+1)!=0We get rid of parentheses
z∈R
2z-1/3z-1+6=0
We multiply all the terms by the denominator
2z*3z-1*3z+6*3z-1=0
Wy multiply elements
6z^2-3z+18z-1=0
We add all the numbers together, and all the variables
6z^2+15z-1=0
a = 6; b = 15; c = -1;
Δ = b2-4ac
Δ = 152-4·6·(-1)
Δ = 249
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-\sqrt{249}}{2*6}=\frac{-15-\sqrt{249}}{12} $$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+\sqrt{249}}{2*6}=\frac{-15+\sqrt{249}}{12} $
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