Question

$$x(4)5m^{2}=22\ m+15$$

Solve for x

$x=\frac{22m+15}{20m^{2}}$
$m\neq 0$

Show Solution

Solve for m (complex solution)

$\left\{\begin{matrix}m=\frac{\sqrt{300x+121}+11}{20x}\text{; }m=\frac{-\sqrt{300x+121}+11}{20x}\text{, }&x\neq 0\\m=-\frac{15}{22}\approx -0.681818182\text{, }&x=0\end{matrix}\right.$

Solve for m

$\left\{\begin{matrix}m=\frac{\sqrt{300x+121}+11}{20x}\text{; }m=\frac{-\sqrt{300x+121}+11}{20x}\text{, }&x\neq 0\text{ and }x\geq -\frac{121}{300}\\m=-\frac{15}{22}\approx -0.681818182\text{, }&x=0\end{matrix}\right.$