Solve (-2-m)x^2-(3m+1)x+4=0

$$( - 2 - m ) x ^ { 2 } - ( 3 m + 1 ) x + 4 = 0$$

Solve for m

$$m=\frac{4-x-2x^{2}}{x\left(x+3\right)}$$

$$x\neq -3\text{ and }x\neq 0$$

Steps for Solving Linear Equation

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

$$-2x^{2}-mx^{2}-\left(3m+1\right)x+4=0$$

Use the distributive property to multiply 3m+1 by x.

$$-2x^{2}-mx^{2}-\left(3mx+x\right)+4=0$$

To find the opposite of 3mx+x, find the opposite of each term.

$$-2x^{2}-mx^{2}-3mx-x+4=0$$

Add 2x^{2} to both sides. Anything plus zero gives itself.

$$-mx^{2}-3mx-x+4=2x^{2}$$

Add x to both sides.

$$-mx^{2}-3mx+4=2x^{2}+x$$

Subtract 4 from both sides.

$$-mx^{2}-3mx=2x^{2}+x-4$$

Combine all terms containing m.

$$\left(-x^{2}-3x\right)m=2x^{2}+x-4$$

Divide both sides by -x^{2}-3x.

$$\frac{\left(-x^{2}-3x\right)m}{-x^{2}-3x}=\frac{2x^{2}+x-4}{-x^{2}-3x}$$

Dividing by -x^{2}-3x undoes the multiplication by -x^{2}-3x.

$$m=\frac{2x^{2}+x-4}{-x^{2}-3x}$$

Divide 2x^{2}+x-4 by -x^{2}-3x.

$$m=\frac{2x^{2}+x-4}{-x\left(x+3\right)}$$

Solve for x

$$\left\{\begin{matrix}x=-\frac{\sqrt{9m^{2}+22m+33}+3m+1}{2\left(m+2\right)}\text{; }x=-\frac{-\sqrt{9m^{2}+22m+33}+3m+1}{2\left(m+2\right)}\text{, }&m\neq -2\\x=-\frac{4}{5}\text{, }&m=-2\end{matrix}\right.$$