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2/3x+2=x-4
We move all terms to the left:
2/3x+2-(x-4)=0
Domain of the equation: 3x!=0We get rid of parentheses
x!=0/3
x!=0
x∈R
2/3x-x+4+2=0
We multiply all the terms by the denominator
-x*3x+4*3x+2*3x+2=0
Wy multiply elements
-3x^2+12x+6x+2=0
We add all the numbers together, and all the variables
-3x^2+18x+2=0
a = -3; b = 18; c = +2;
Δ = b2-4ac
Δ = 182-4·(-3)·2
Δ = 348
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{348}=\sqrt{4*87}=\sqrt{4}*\sqrt{87}=2\sqrt{87}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(18)-2\sqrt{87}}{2*-3}=\frac{-18-2\sqrt{87}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(18)+2\sqrt{87}}{2*-3}=\frac{-18+2\sqrt{87}}{-6} $
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