Calculate $\sqrt{x+6}$ to the power of $2$ and get $x+6$.
$$x+6=x^{2}$$
Subtract $x^{2}$ from both sides.
$$x+6-x^{2}=0$$
Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.
$$-x^{2}+x+6=0$$
To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as $-x^{2}+ax+bx+6$. To find $a$ and $b$, set up a system to be solved.
$$a+b=1$$ $$ab=-6=-6$$
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. List all such integer pairs that give product $-6$.
$$-1,6$$ $$-2,3$$
Calculate the sum for each pair.
$$-1+6=5$$ $$-2+3=1$$
The solution is the pair that gives sum $1$.
$$a=3$$ $$b=-2$$
Rewrite $-x^{2}+x+6$ as $\left(-x^{2}+3x\right)+\left(-2x+6\right)$.
$$\left(-x^{2}+3x\right)+\left(-2x+6\right)$$
Factor out $-x$ in the first and $-2$ in the second group.
$$-x\left(x-3\right)-2\left(x-3\right)$$
Factor out common term $x-3$ by using distributive property.
$$\left(x-3\right)\left(-x-2\right)$$
To find equation solutions, solve $x-3=0$ and $-x-2=0$.
$$x=3$$ $$x=-2$$
Substitute $3$ for $x$ in the equation $\sqrt{x+6}=x$.
$$\sqrt{3+6}=3$$
Simplify. The value $x=3$ satisfies the equation.
$$3=3$$
Substitute $-2$ for $x$ in the equation $\sqrt{x+6}=x$.
$$\sqrt{-2+6}=-2$$
Simplify. The value $x=-2$ does not satisfy the equation because the left and the right hand side have opposite signs.
