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

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Solution for (x^2+2)^1/2-(2x+5)^1/2=0 equation: 


x in (-oo:+oo)

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

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

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

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

x^2-2*x-3 = 0

x^2-2*x-3 = 0

x^2-2*x-3 = 0

DELTA = (-2)^2-(-3*1*4)

DELTA = 16

DELTA > 0

x = (16^(1/2)+2)/(1*2) or x = (2-16^(1/2))/(1*2)

x = 3 or x = -1

(x+1)*(x-3) = 0

((x+1)*(x-3))/2 = 0

((x+1)*(x-3))/2 = 0 // * 2

(x+1)*(x-3) = 0

( x+1 )

x+1 = 0 // - 1

x = -1

( x-3 )

x-3 = 0 // + 3

x = 3

x in { -1, 3 }

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