Solve left(m+1right)x^2-2mx+m-5=0

$$\left( m+1 \right) { x }^{ 2 } -2mx+m-5 = 0$$

Solve for m

$$m=-\frac{x^{2}-5}{\left(x-1\right)^{2}}$$

$$x\neq 1$$

Steps for Solving Linear Equation

Use the distributive property to multiply m+1 by x^{2}.

$$mx^{2}+x^{2}-2mx+m-5=0$$

Subtract x^{2} from both sides. Anything subtracted from zero gives its negation.

$$mx^{2}-2mx+m-5=-x^{2}$$

Add 5 to both sides.

$$mx^{2}-2mx+m=-x^{2}+5$$

Combine all terms containing m.

$$\left(x^{2}-2x+1\right)m=-x^{2}+5$$

The equation is in standard form.

$$\left(x^{2}-2x+1\right)m=5-x^{2}$$

Divide both sides by x^{2}-2x+1.

$$\frac{\left(x^{2}-2x+1\right)m}{x^{2}-2x+1}=\frac{5-x^{2}}{x^{2}-2x+1}$$

Dividing by x^{2}-2x+1 undoes the multiplication by x^{2}-2x+1.

$$m=\frac{5-x^{2}}{x^{2}-2x+1}$$

Divide -x^{2}+5 by x^{2}-2x+1.

$$m=\frac{5-x^{2}}{\left(x-1\right)^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{m+\sqrt{4m+5}}{m+1}\text{; }x=\frac{m-\sqrt{4m+5}}{m+1}\text{, }&m\neq -1\\x=3\text{, }&m=-1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{m+\sqrt{4m+5}}{m+1}\text{; }x=\frac{m-\sqrt{4m+5}}{m+1}\text{, }&m\neq -1\text{ and }m\geq -\frac{5}{4}\\x=3\text{, }&m=-1\end{matrix}\right.$$