Let $p = \frac{1}{2} \sqrt{\dfrac{x^2 - 3y}{x^2}}$

1. Substitute $x=-12$ and $y=4$ into the formula and evaluate.

$p=\frac12\sqrt{\frac{x^2-3y}{x^2}}$
$p=\frac12\sqrt{\frac{(-12)^2-3(4)}{(-12)^2}}$
$=\frac12\sqrt{\frac{144-12}{144}}$
$=\frac12\sqrt{\frac{132}{144}}$
$=0.4787\ldots$
Correct to two decimal places,
$p=0.48$

1. Remove the coefficient $\frac12$ and square both sides to remove the square root.

$p=\dfrac12\sqrt{\dfrac{x^2-3y}{x^2}}$

$2p=\sqrt{\dfrac{x^2-3y}{x^2}}$

$(2p)^2=\dfrac{x^2-3y}{x^2}$

$4p^2=\dfrac{x^2-3y}{x^2}$

2. Clear the denominator involving $x^2$.

$4p^2=\dfrac{x^2-3y}{x^2}$

$4p^2x^2=x^2-3y$

3. Collect the $x^2$ terms and factorise.

$4p^2x^2=x^2-3y$

$x^2-4p^2x^2=3y$

$x^2(1-4p^2)=3y$

4. Isolate $x^2$ and take the square root to make $x$ the subject.

$x^2(1-4p^2)=3y$

$x^2=\dfrac{3y}{1-4p^2}$

$x=\pm\sqrt{\dfrac{3y}{1-4p^2}}$