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2/5x-1/2x=40
We move all terms to the left:
2/5x-1/2x-(40)=0
Domain of the equation: 5x!=0
x!=0/5
x!=0
x∈R
Domain of the equation: 2x!=0We calculate fractions
x!=0/2
x!=0
x∈R
4x/10x^2+(-5x)/10x^2-40=0
We multiply all the terms by the denominator
4x+(-5x)-40*10x^2=0
Wy multiply elements
-400x^2+4x+(-5x)=0
We get rid of parentheses
-400x^2+4x-5x=0
We add all the numbers together, and all the variables
-400x^2-1x=0
a = -400; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·(-400)·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$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*-400}=\frac{0}{-800} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*-400}=\frac{2}{-800} =-1/400 $
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