Question Number 89153 by Jidda28 last updated on 15/Apr/20
$$\mathrm{If}\:\mathrm{P}=\frac{\mathrm{RE}^{\mathrm{2}} }{\left(\mathrm{R}+\mathrm{B}\right)^{\mathrm{2}} }\:\mathrm{make}\:\mathrm{R}\:\mathrm{the}\:\mathrm{subject}\:\mathrm{of}\:\mathrm{the}\:\mathrm{formula}. \\ $$
Answered by MJS last updated on 15/Apr/20
$${p}\left({r}+{b}\right)^{\mathrm{2}} ={re}^{\mathrm{2}} \\ $$$$\left({r}+{b}\right)^{\mathrm{2}} =\frac{{e}^{\mathrm{2}} }{{p}}{r} \\ $$$${r}^{\mathrm{2}} +\left(\mathrm{2}{b}−\frac{{e}^{\mathrm{2}} }{{p}}\right){r}+{b}^{\mathrm{2}} =\mathrm{0} \\ $$$${r}=\frac{{e}^{\mathrm{2}} }{\mathrm{2}{p}}−{b}\pm\sqrt{\left({b}−\frac{{e}^{\mathrm{2}} }{\mathrm{2}{p}}\right)^{\mathrm{2}} −{b}^{\mathrm{2}} } \\ $$$${r}=\frac{{e}^{\mathrm{2}} }{\mathrm{2}{p}}−{b}\pm\frac{{e}}{\mathrm{2}{p}}\sqrt{{e}^{\mathrm{2}} −\mathrm{4}{bp}} \\ $$
Commented by Jidda28 last updated on 15/Apr/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$