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2/x+3=5/4x
We move all terms to the left:
2/x+3-(5/4x)=0
Domain of the equation: x!=0
x∈R
Domain of the equation: 4x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
2/x-(+5/4x)+3=0
We get rid of parentheses
2/x-5/4x+3=0
We calculate fractions
8x/4x^2+(-5x)/4x^2+3=0
We multiply all the terms by the denominator
8x+(-5x)+3*4x^2=0
Wy multiply elements
12x^2+8x+(-5x)=0
We get rid of parentheses
12x^2+8x-5x=0
We add all the numbers together, and all the variables
12x^2+3x=0
a = 12; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·12·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*12}=\frac{-6}{24} =-1/4 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*12}=\frac{0}{24} =0 $
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