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5/2x+3=1/x
We move all terms to the left:
5/2x+3-(1/x)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
5/2x-(+1/x)+3=0
We get rid of parentheses
5/2x-1/x+3=0
We calculate fractions
5x/2x^2+(-2x)/2x^2+3=0
We multiply all the terms by the denominator
5x+(-2x)+3*2x^2=0
Wy multiply elements
6x^2+5x+(-2x)=0
We get rid of parentheses
6x^2+5x-2x=0
We add all the numbers together, and all the variables
6x^2+3x=0
a = 6; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·6·0
Δ = 9
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}$$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$$\sqrt{\Delta}=\sqrt{9}=3$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*6}=\frac{-6}{12} =-1/2 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*6}=\frac{0}{12} =0 $
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