x+5=(2x-7)^1/2

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Solution for x+5=(2x-7)^1/2 equation: 


x in (-oo:+oo)

x+5 = ((2*x-7)^1)/2 // - ((2*x-7)^1)/2

x-(((2*x-7)^1)/2)+5 = 0

x-((2*x-7)/2)+5 = 0

(-1*(2*x-7))/2+x+5 = 0

(-1*(2*x-7))/2+(2*x)/2+(2*5)/2 = 0

2*x-1*(2*x-7)+2*5 = 0

7+10 = 0

17 = 0

17/2 = 0

17/2 = 0 // * 2

17 = 0

x belongs to the empty set

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