We have to solve:
solve x2−10x=−21

\(\displaystyle{x}^{{2}}-{10}{x}=-{21}\to{x}^{{2}}-{10}{x}+{21}={0}\)

For a quadratic equation of the form \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}\) the solution are \(\displaystyle\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{2}}{a}\)

Therefore the solution of the given equation are \(\displaystyle\frac{{-{\left({10}\right)}\pm\sqrt{{{\left(-{10}\right)}^{{2}}}}-{4}\cdot{1}\cdot{21}}}{{2}}\cdot{1}=\frac{{{10}+\sqrt{{16}}}}{{2}}\cdot{1}=\frac{{{10}\pm{4}}}{{2}}={7.3}\)

Hence the solution are x=7 and x=3

\(\displaystyle{x}^{{2}}-{10}{x}=-{21}\to{x}^{{2}}-{10}{x}+{21}={0}\)

For a quadratic equation of the form \(\displaystyle{a}{x}^{{2}}+{b}{x}+{c}={0}\) the solution are \(\displaystyle\frac{{-{b}\pm\sqrt{{{b}^{{2}}-{4}{a}{c}}}}}{{2}}{a}\)

Therefore the solution of the given equation are \(\displaystyle\frac{{-{\left({10}\right)}\pm\sqrt{{{\left(-{10}\right)}^{{2}}}}-{4}\cdot{1}\cdot{21}}}{{2}}\cdot{1}=\frac{{{10}+\sqrt{{16}}}}{{2}}\cdot{1}=\frac{{{10}\pm{4}}}{{2}}={7.3}\)

Hence the solution are x=7 and x=3