Solve the following question in detail:
How do you solve \[ - 3{x^2} + 12 = 0\]?

Hint: Take the given equation and divide it to reduce the coefficient magnitude. Then use the standard equation of the quadratic form for the two-degree equation to deduce the exact values of the coefficients. Then we substitute these values in the quadratic formula to get the final values of the variable.

Complete step-by-step solution:
Given equation;
\[ - 3{x^2} + 12 = 0\]
Let us simplify the equation by dividing both sides by \[ - 3\]to reduce the coefficients while still keeping the equation balanced.
\[ \Rightarrow \dfrac{{ - 3{x^2} + 12}}{{ - 3}} = \dfrac{0}{{ - 3}}\]
Now, we can separate each and every term by the numerator and denominator. We get;
\[ \Rightarrow \dfrac{{ - 3{x^2}}}{{ - 3}} + \dfrac{{12}}{{ - 3}} = 0\]
We reduce the equation by dividing the numerator with the denominator. We get;
\[ \Rightarrow {x^2} - 4 = 0\]
Firstly, the most efficient method to find the value of the variable, we have the quadratic formula.
\[x = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}\]
Here, \[a,b,c\]are the coefficients of the variables for each degree sorted in ascending order respectively.
 The standard quadratic equation formula for a two-degree equation is: \[a{x^2} + bx + c = 0\]
Here, we compare the acquired equation to the standard quadratic equation to find the values of the coefficients.
\[a = 1\]
\[b = 0\]
\[c = - 4\]
Now, substituting these values in the formula of the quadratic equations, we get;
\[x = \dfrac{{ - 0 \pm \sqrt {{0^2} - 4 \times 1\left( { - 4} \right)} }}{{2 \times 1}}\]
Simplifying the equation by multiplying all the terms, we get;
$\Rightarrow$\[x = \dfrac{{ \pm \sqrt {0 + 16} }}{2}\]
Now, by adding the terms, we get;
$\Rightarrow$\[x = \dfrac{{ \pm \sqrt {16} }}{2}\]
Taking the value in the root out we get a perfect square. That means, now, we have;
$\Rightarrow$\[x = \dfrac{{ \pm 4}}{2}\]
Now, we divide the numerator with the denominator. We get;
$\Rightarrow$\[x = \pm 2\]
Now, once we take the positive value and next, we take the negative value. Taking the positive value, we get;
$\Rightarrow$\[x = 2\]
Taking the negative value, we get;
$\Rightarrow$\[x = - 2\]
So, we have the value for \[x\]

Therefore, \[x = \pm 2\]

Note: we can solve this equation by a simpler method. We have,
\[{x^2} - 4 = 0\]
By taking the constant on the other side, we get;
\[ \Rightarrow {x^2} = 4\]
Now, applying square root on both the sides, we get;
\[ \Rightarrow \sqrt {{x^2}} = \sqrt 4 \]
Taking the terms out of square root, we get;
\[ \Rightarrow x = \pm 2\]