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2/2x-8=5/x+4
We move all terms to the left:
2/2x-8-(5/x+4)=0
Domain of the equation: 2x!=0
x!=0/2
x!=0
x∈R
Domain of the equation: x+4)!=0We get rid of parentheses
x∈R
2/2x-5/x-4-8=0
We calculate fractions
2x/2x^2+(-10x)/2x^2-4-8=0
We add all the numbers together, and all the variables
2x/2x^2+(-10x)/2x^2-12=0
We multiply all the terms by the denominator
2x+(-10x)-12*2x^2=0
Wy multiply elements
-24x^2+2x+(-10x)=0
We get rid of parentheses
-24x^2+2x-10x=0
We add all the numbers together, and all the variables
-24x^2-8x=0
a = -24; b = -8; c = 0;
Δ = b2-4ac
Δ = -82-4·(-24)·0
Δ = 64
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{64}=8$$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-8}{2*-24}=\frac{0}{-48} =0 $$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+8}{2*-24}=\frac{16}{-48} =-1/3 $
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