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(1)/(3)x+4=x-2
We move all terms to the left:
(1)/(3)x+4-(x-2)=0
Domain of the equation: 3x!=0We get rid of parentheses
x!=0/3
x!=0
x∈R
1/3x-x+2+4=0
We multiply all the terms by the denominator
-x*3x+2*3x+4*3x+1=0
Wy multiply elements
-3x^2+6x+12x+1=0
We add all the numbers together, and all the variables
-3x^2+18x+1=0
a = -3; b = 18; c = +1;
Δ = b2-4ac
Δ = 182-4·(-3)·1
Δ = 336
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{336}=\sqrt{16*21}=\sqrt{16}*\sqrt{21}=4\sqrt{21}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-4\sqrt{21}}{2*-3}=\frac{-18-4\sqrt{21}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+4\sqrt{21}}{2*-3}=\frac{-18+4\sqrt{21}}{-6} $
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