# Find the solution: 6=\sqrt{x^2-2x+12}

Find the solution:
$$\displaystyle{6}=\sqrt{{{x}^{{2}}-{2}{x}+{12}}}$$

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Jayden-James Duffy

Radical or Exponents or Power is defined as the number of times that a variable or constant is being repeated. It is also called power raised. Exponents are used to describing number of times in a single notation
The radical equation given to us
$$\displaystyle{6}=\sqrt{{{x}^{{2}}-{2}{x}+{12}}}$$
To find the solution for the given radical equation
Solution to the question
We have,
$$\displaystyle{6}=\sqrt{{{x}^{{2}}-{2}{x}+{12}}}$$
Squaring both sides we obtain,
$$\displaystyle{6}^{{2}}={\left(\sqrt{{{x}^{{2}}-{2}{x}+{12}}}\right)}^{{2}}$$
$$\displaystyle{36}={x}^{{2}}-{2}{x}+{12}$$
$$\displaystyle{x}^{{2}}-{2}{x}+{12}-{36}={0}$$
$$\displaystyle{x}^{{2}}-{2}{x}-{24}={0}$$
$$\displaystyle{x}^{{2}}-{6}{x}+{4}{x}={0}$$
$$\displaystyle{x}{\left({x}-{6}\right)}+{4}{\left({x}-{6}\right)}={0}$$
$$\displaystyle{\left({x}+{4}\right)}{\left({x}-{6}\right)}={0}$$
either $$x=-4\ \text{or}\ x=6$$
There are two valid solutions $$\displaystyle{x}=-{4}{\quad\text{and}\quad}{x}={6}$$