sqrt(x+3)=x-3 equation

Given the equation
$$\sqrt{x + 3} = x - 3$$
$$\sqrt{x + 3} = x - 3$$
We raise the equation sides to 2-th degree
$$x + 3 = \left(x - 3\right)^{2}$$
$$x + 3 = x^{2} - 6 x + 9$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 7 x - 6 = 0$$
This equation is of the form
$$a\ x^2 + b\ x + c = 0$$
A quadratic equation can be solved using the discriminant
The roots of the quadratic equation:
$$x_{1} = \frac{\sqrt{D} - b}{2 a}$$
$$x_{2} = \frac{- \sqrt{D} - b}{2 a}$$
where $D = b^2 - 4 a c$ is the discriminant.
Because
$$a = -1$$
$$b = 7$$
$$c = -6$$
, then
$$D = b^2 - 4\ a\ c = $$
$$\left(-1\right) \left(\left(-1\right) 4\right) \left(-6\right) + 7^{2} = 25$$
Because D > 0, then the equation has two roots.
$$x_1 = \frac{(-b + \sqrt{D})}{2 a}$$
$$x_2 = \frac{(-b - \sqrt{D})}{2 a}$$
or
$$x_{1} = 1$$
Simplify
$$x_{2} = 6$$
Simplify

Because
$$\sqrt{x + 3} = x - 3$$
and
$$\sqrt{x + 3} \geq 0$$
then
$$x - 3 >= 0$$
or
$$3 \leq x$$
$$x < \infty$$
The final answer:
$$x_{2} = 6$$