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 \(\displaystyle{\left\lbrace{3}\right\rbrace}\).

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 \(\displaystyle{\left\lbrace{3}\right\rbrace}\).