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x-3/3x+1=-3
We move all terms to the left:
x-3/3x+1-(-3)=0
Domain of the equation: 3x!=0We add all the numbers together, and all the variables
x!=0/3
x!=0
x∈R
x-3/3x+4=0
We multiply all the terms by the denominator
x*3x+4*3x-3=0
Wy multiply elements
3x^2+12x-3=0
a = 3; b = 12; c = -3;
Δ = b2-4ac
Δ = 122-4·3·(-3)
Δ = 180
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{180}=\sqrt{36*5}=\sqrt{36}*\sqrt{5}=6\sqrt{5}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(12)-6\sqrt{5}}{2*3}=\frac{-12-6\sqrt{5}}{6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(12)+6\sqrt{5}}{2*3}=\frac{-12+6\sqrt{5}}{6} $
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