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10-x+2=1/3x+4
We move all terms to the left:
10-x+2-(1/3x+4)=0
Domain of the equation: 3x+4)!=0We add all the numbers together, and all the variables
x∈R
-1x-(1/3x+4)+12=0
We get rid of parentheses
-1x-1/3x-4+12=0
We multiply all the terms by the denominator
-1x*3x-4*3x+12*3x-1=0
Wy multiply elements
-3x^2-12x+36x-1=0
We add all the numbers together, and all the variables
-3x^2+24x-1=0
a = -3; b = 24; c = -1;
Δ = b2-4ac
Δ = 242-4·(-3)·(-1)
Δ = 564
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{564}=\sqrt{4*141}=\sqrt{4}*\sqrt{141}=2\sqrt{141}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(24)-2\sqrt{141}}{2*-3}=\frac{-24-2\sqrt{141}}{-6} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(24)+2\sqrt{141}}{2*-3}=\frac{-24+2\sqrt{141}}{-6} $
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