91Ó°ÊÓ

The equation given to us is \( (m-6)^{2}=20 \). To solve it, we use the square root property. When an equation is in the form of \( x^2 = a \, \), we can solve for \ x \ by taking the square root of both sides.

This gives us: \( x = \pm\rightarrow \sqrt{a} \). We must consider both the positive and negative roots because squaring either will return the original positive value.

Applying this to our problem: \[ \sqrt{ (m-6)^{2} }= \pm\rightarrow \sqrt{20} \]

Notice this gives us: \[ m-6 = \pm\rightarrow \sqrt{20} \]
Which simplifies further to:
\[ m-6 = \pm\rightarrow 2\sqrt{5} \]

The square root property is useful when solving quadratic equations, especially when the equation is already in a square form, saving quite a bit of time.

In earlier steps, we've isolated the squared term, \ (m-6)^2 \ on one side of the equation, making it easier to apply the square root property.

Now, we need to isolate the variable \ m \.

After taking the square root of both sides, we have: \[ m-6 = \pm \rightarrow 2\sqrt{5} \]

To isolate \ m \ from here, we need to remove the \ -6\term attached to \ m by performing the opposite operation. We add \ 6 \ to both sides of the equation:

\[ m -6 + 6 = 6 \pm \rightarrow 2\sqrt{5 } \] Which simplifies nicely to: \[ m= 6 \pm \rightarrow 2\sqrt{5} \]

Always remember, isolating the variable step ensures that we're left with \ m \ on one side and the resolved values or expressions on the other side of the equation.

Now, we have the expression \ m= 6 \pm \rightarrow 2\sqrt{5} \ from our previous steps. This means there are two potential solutions for \ m.

Let's break it down:

\[ m = 6 + 2\sqrt{5} \]
and
\[ m = 6 - 2\sqrt{5} \]

These show the two values \ m \ can take.

Solving for \ m \ involves finding the exact numerical values, but knowing it is adequate to leave the answer in simplified radical form unless specified to find a decimal approximation. However, if required, you could use a calculator for more precise values:
- For \ 6 + 2\sqrt{5} \
- For \ 6 - 2\sqrt{5} \

Remember, these types of problems help us appreciate the power of algebra in solving for variables.