# To calculate: The solution of the equation x^{2}-5=\left(x+2\right)

To calculate: The solution of the equation $$\displaystyle{x}^{{{2}}}-{5}={\left({x}+{2}\right)}{\left({x}-{4}\right)}$$.

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Formula used:
The distributive property is $$\displaystyle{a}{\left({b}+{c}\right)}={a}{b}+{a}{c}$$.
Calculation:
Consider the equation $$\displaystyle{x}^{{{2}}}-{5}={\left({x}+{2}\right)}{\left({x}-{4}\right)}$$.
Apply the distributive property.
$$\displaystyle{x}^{{{2}}}-{5}={x}^{{{2}}}-{4}{x}+{2}{x}-{8}$$
$$\displaystyle{x}^{{{2}}}-{5}={x}^{{{2}}}-{2}{x}-{8}$$
$$\displaystyle-{5}=-{2}{x}-{8}$$
$$\displaystyle{8}-{5}=-{2}{x}$$
Further solve the equation.
$$\displaystyle{3}=-{2}{x}$$
$$\displaystyle{x}=-{\frac{{{3}}}{{{2}}}}$$
Check the solution $$\displaystyle{x}=-{\frac{{{3}}}{{{2}}}}$$ in the equation $$\displaystyle{x}^{{{2}}}-{5}={\left({x}+{2}\right)}{\left({x}-{4}\right)}$$.
$$\displaystyle{\left(-{\frac{{{3}}}{{{2}}}}\right)}^{{{2}}}-{5}{\overset{{?}}{{=}}}{\left\lbrace{\left(-{\frac{{{3}}}{{{2}}}}\right)}+{2}\right\rbrace}{\left\lbrace{\left(-{\frac{{{3}}}{{{2}}}}\right)}-{4}\right\rbrace}$$
$$\displaystyle{\frac{{{9}}}{{{4}}}}-{5}{\overset{{?}}{{=}}}{\left({\frac{{-{3}+{4}}}{{{2}}}}\right)}{\left({\frac{{-{3}-{8}}}{{{2}}}}\right)}$$
$$\displaystyle{\frac{{{9}-{20}}}{{{4}}}}{\overset{{?}}{{=}}}{\left({\frac{{{1}}}{{{2}}}}\right)}{\left(-{\frac{{{11}}}{{{2}}}}\right)}$$
$$\displaystyle-{\frac{{{11}}}{{{4}}}}=-{\frac{{{11}}}{{{4}}}}$$
Thus, this solution is true.
Therefore, thesolution of the equation $$\displaystyle{x}^{{{2}}}-{5}={\left({x}+{2}\right)}{\left({x}-{4}\right)}$$ is $$\displaystyle{\left\lbrace{3}\right\rbrace}$$.