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5/2+3/20x=2/5x
We move all terms to the left:
5/2+3/20x-(2/5x)=0
Domain of the equation: 20x!=0
x!=0/20
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
3/20x-(+2/5x)+5/2=0
We get rid of parentheses
3/20x-2/5x+5/2=0
We calculate fractions
2500x^2/400x^2+60x/400x^2+(-160x)/400x^2=0
We multiply all the terms by the denominator
2500x^2+60x+(-160x)=0
We get rid of parentheses
2500x^2+60x-160x=0
We add all the numbers together, and all the variables
2500x^2-100x=0
a = 2500; b = -100; c = 0;
Δ = b2-4ac
Δ = -1002-4·2500·0
Δ = 10000
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{10000}=100$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-100)-100}{2*2500}=\frac{0}{5000} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-100)+100}{2*2500}=\frac{200}{5000} =1/25 $
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