Please use English while explaining

The expressions under the square roots must be non-negative:
$4-x \ge 0 \implies x \le 4$
$6+x \ge 0 \implies x \ge -6$
$14+2x \ge 0 \implies 2x \ge -14 \implies x \ge -7$
Therefore, $x$ must be in the interval $[-6, 4]$.

Rearrange the equation to isolate one radical:

Square both sides:

Divide both sides by 2:

Since the right side is a square root (and thus non-negative), the left side must also be non-negative:

So, any solution must satisfy the initial domain conditions and $x \le -4$. This means $x$ must be in the interval $[-6, -4]$.

Divide by 2:

Factor the quadratic equation:

Thus, $x=-5$ or $x=2$.

From Step 3, we required $x \in [-6, -4]$.

• For $x=-5$: $-5$ is within the interval $[-6, -4]$.
• For $x=2$: 2 is not within the interval $[-6, -4]$. Thus, $x=2$ is an extraneous solution.

For $x=-5$:

The solution is correct.

The solution for equation a) is: $\boxed{x=-5}$

The expressions under the square roots must be non-negative:
$x-12 \ge 0 \implies x \ge 12$
$2x-3 \ge 0 \implies 2x \ge 3 \implies x \ge \frac{3}{2}$
Therefore, $x$ must be in the interval $[12, \infty)$.

Rearrange the equation:

Square both sides:

Divide both sides by 2:

Since the left side (square root) is non-negative, the right side must also be non-negative:

From Step 1, we required $x \ge 12$.
From Step 3, we required $x \le 10$.
There is no value of $x$ that can satisfy both conditions ($x \ge 12$ and $x \le 10$) simultaneously. This means there is no valid solution for this equation.

The solution for equation b) is: $\boxed{\text{No solution}}$