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10/3x=3x-2
We move all terms to the left:
10/3x-(3x-2)=0
Domain of the equation: 3x!=0We get rid of parentheses
x!=0/3
x!=0
x∈R
10/3x-3x+2=0
We multiply all the terms by the denominator
-3x*3x+2*3x+10=0
Wy multiply elements
-9x^2+6x+10=0
a = -9; b = 6; c = +10;
Δ = b2-4ac
Δ = 62-4·(-9)·10
Δ = 396
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$
The end solution:
$\sqrt{\Delta}=\sqrt{396}=\sqrt{36*11}=\sqrt{36}*\sqrt{11}=6\sqrt{11}$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-6\sqrt{11}}{2*-9}=\frac{-6-6\sqrt{11}}{-18} $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+6\sqrt{11}}{2*-9}=\frac{-6+6\sqrt{11}}{-18} $
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