sqrt(16-x^2)=2 equation

Given the equation
$$\sqrt{- x^{2} + 16} = 2$$
$$\sqrt{- x^{2} + 16} = 2$$
We raise the equation sides to 2-th degree
$$- x^{2} + 16 = 4$$
$$- x^{2} + 16 = 4$$
Transfer the right side of the equation left part with negative sign
$$- x^{2} + 12 = 0$$
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
$$a = -1$$
$$b = 0$$
$$c = 12$$
, then
$$D = b^2 - 4\ a\ c = $$
$$0^{2} - \left(-1\right) 4 \cdot 12 = 48$$
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
$$x_{1} = - 2 \sqrt{3}$$
Simplify
$$x_{2} = 2 \sqrt{3}$$
Simplify

Because
$$\sqrt{- x^{2} + 16} = 2$$
and
$$\sqrt{- x^{2} + 16} \geq 0$$
then
$$2 >= 0$$
The final answer:
$$x_{1} = - 2 \sqrt{3}$$
$$x_{2} = 2 \sqrt{3}$$