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2/5x+3=1/4x
We move all terms to the left:
2/5x+3-(1/4x)=0
Domain of the equation: 5x!=0
x!=0/5
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
2/5x-(+1/4x)+3=0
We get rid of parentheses
2/5x-1/4x+3=0
We calculate fractions
8x/20x^2+(-5x)/20x^2+3=0
We multiply all the terms by the denominator
8x+(-5x)+3*20x^2=0
Wy multiply elements
60x^2+8x+(-5x)=0
We get rid of parentheses
60x^2+8x-5x=0
We add all the numbers together, and all the variables
60x^2+3x=0
a = 60; b = 3; c = 0;
Δ = b2-4ac
Δ = 32-4·60·0
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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(3)-3}{2*60}=\frac{-6}{120} =-1/20 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(3)+3}{2*60}=\frac{0}{120} =0 $
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