Solve for X: Finding Solutions to x²/10 - 10 = 0

Let's solve the given equation:

$\frac{x^2}{10}-10=0$

We'll start by getting rid of the fraction line in the left side of the given equation, we'll do this by multiplying both sides of the equation by the common denominator - which is the number 10, then we'll move the free number to one side, remembering that when moving a term between sides - its sign changes:

$\frac{x^2}{10}-\frac{10}{1}=0\hspace{8pt}\text{/}\cdot 10\\ \\ 1\cdot x^2-10\cdot10=0 \\ x^2-100=0\\ x^2=100$

From here, we'll solve simply, we'll perform on both sides the opposite operation to the square power operation applied to the unknown in the equation, which is the square root operation, using the laws of exponents:

a. Definition of root as a power:

$\sqrt[n]{a}=a^{\frac{1}{n}}$and the two laws of exponents:

b. Law of exponents for power of a power:

$(a^m)^n=a^{m\cdot n}$

Let's continue solving the equation:
$x^2=100\hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ \sqrt{ x^2}=\pm\sqrt{ 100}\\ (x^2)^{\frac{1}{2}}=\pm10\\ x^{2\cdot\frac{1}{2}}=\pm10\\ \boxed{x=10,-10}$

In the first stage, we applied the square root to both sides of the equation, then we remembered the definition of root as a power (a.) on the left side, in the next stage - we applied the law of exponents for power of a power (b.) on the left side, and remembered that raising a number to the power of 1 doesn't change the number.

Additionally, we remembered that since an even power doesn't preserve the sign of the number it's applied to (will always give a positive result), taking an even root of both sides of the equation requires considering two possible cases - positive and negative (this is unlike taking a root of an odd order, which requires considering only one case that matches the sign of the number the root is applied to),

Let's summarize the solution of the equation:

$\frac{x^2}{10}-10=0 \hspace{8pt}\text{/}\cdot 10\\ x^2=100 \hspace{8pt}\text{/}\sqrt{\hspace{6pt}}\\ \boxed{x=10,-10}$

Therefore, the correct answer is answer a.