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

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

Solve for m

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

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

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.

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

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

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

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 (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(m-1\right)\left(9m+31\right)}+3m+1}{2\left(2-m\right)}\text{; }x=\frac{-\sqrt{\left(m-1\right)\left(9m+31\right)}+3m+1}{2\left(2-m\right)}\text{, }&m\neq 2\\x=\frac{4}{7}\text{, }&m=2\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(m-1\right)\left(9m+31\right)}+3m+1}{2\left(2-m\right)}\text{; }x=\frac{-\sqrt{\left(m-1\right)\left(9m+31\right)}+3m+1}{2\left(2-m\right)}\text{, }&\left(m\neq 2\text{ and }m\geq 1\right)\text{ or }m\leq -\frac{31}{9}\\x=\frac{4}{7}\text{, }&m=2\end{matrix}\right.$$