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2/5x+3=1/2x+5
We move all terms to the left:
2/5x+3-(1/2x+5)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 2x+5)!=0We get rid of parentheses
x∈R
2/5x-1/2x-5+3=0
We calculate fractions
4x/10x^2+(-5x)/10x^2-5+3=0
We add all the numbers together, and all the variables
4x/10x^2+(-5x)/10x^2-2=0
We multiply all the terms by the denominator
4x+(-5x)-2*10x^2=0
Wy multiply elements
-20x^2+4x+(-5x)=0
We get rid of parentheses
-20x^2+4x-5x=0
We add all the numbers together, and all the variables
-20x^2-1x=0
a = -20; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-20)·0
Δ = 1
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{1}=1$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-20}=\frac{0}{-40} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-20}=\frac{2}{-40} =-1/20 $
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