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1/4x+28=3/5x
We move all terms to the left:
1/4x+28-(3/5x)=0
Domain of the equation: 4x!=0
x!=0/4
x!=0
x∈R
Domain of the equation: 5x)!=0We add all the numbers together, and all the variables
x!=0/1
x!=0
x∈R
1/4x-(+3/5x)+28=0
We get rid of parentheses
1/4x-3/5x+28=0
We calculate fractions
5x/20x^2+(-12x)/20x^2+28=0
We multiply all the terms by the denominator
5x+(-12x)+28*20x^2=0
Wy multiply elements
560x^2+5x+(-12x)=0
We get rid of parentheses
560x^2+5x-12x=0
We add all the numbers together, and all the variables
560x^2-7x=0
a = 560; b = -7; c = 0;
Δ = b2-4ac
Δ = -72-4·560·0
Δ = 49
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{49}=7$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-7)-7}{2*560}=\frac{0}{1120} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-7)+7}{2*560}=\frac{14}{1120} =1/80 $
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