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\left(\sqrt{x+2}\right)^{2}=x^{2}
Square both sides of the equation.
x+2=x^{2}
Calculate \sqrt{x+2} to the power of 2 and get x+2.
x+2-x^{2}=0
Subtract x^{2} from both sides.
-x^{2}+x+2=0
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
a+b=1 ab=-2=-2
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -x^{2}+ax+bx+2. To find a and b, set up a system to be solved.
a=2 b=-1
Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.
\left(-x^{2}+2x\right)+\left(-x+2\right)
Rewrite -x^{2}+x+2 as \left(-x^{2}+2x\right)+\left(-x+2\right).
-x\left(x-2\right)-\left(x-2\right)
Factor out -x in the first and -1 in the second group.
\left(x-2\right)\left(-x-1\right)
Factor out common term x-2 by using distributive property.
x=2 x=-1
To find equation solutions, solve x-2=0 and -x-1=0.
\sqrt{2+2}=2
Substitute 2 for x in the equation \sqrt{x+2}=x.
2=2
Simplify. The value x=2 satisfies the equation.
\sqrt{-1+2}=-1
Substitute -1 for x in the equation \sqrt{x+2}=x.
1=-1
Simplify. The value x=-1 does not satisfy the equation because the left and the right hand side have opposite signs.
x=2
Equation \sqrt{x+2}=x has a unique solution.