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

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

Solve for m

$$m=\frac{3x^{2}+1}{2x^{2}+4x+1}$$

$$x\neq \frac{\sqrt{2}}{2}-1\text{ and }x\neq -\frac{\sqrt{2}}{2}-1$$

Steps for Solving Linear Equation

Use the distributive property to multiply 2m-3 by x^{2}.

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

Add 3x^{2} to both sides. Anything plus zero gives itself.

$$2mx^{2}+4mx+m-1=3x^{2}$$

Add 1 to both sides.

$$2mx^{2}+4mx+m=3x^{2}+1$$

Combine all terms containing m.

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

Divide both sides by 2x^{2}+4x+1.

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

Dividing by 2x^{2}+4x+1 undoes the multiplication by 2x^{2}+4x+1.

$$m=\frac{3x^{2}+1}{2x^{2}+4x+1}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(2m-1\right)\left(m+3\right)}-2m}{2m-3}\text{; }x=-\frac{\sqrt{\left(2m-1\right)\left(m+3\right)}+2m}{2m-3}\text{, }&m\neq \frac{3}{2}\\x=-\frac{1}{12}\text{, }&m=\frac{3}{2}\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{\left(2m-1\right)\left(m+3\right)}-2m}{2m-3}\text{; }x=-\frac{\sqrt{\left(2m-1\right)\left(m+3\right)}+2m}{2m-3}\text{, }&\left(m\neq \frac{3}{2}\text{ and }m\geq \frac{1}{2}\right)\text{ or }m\leq -3\\x=-\frac{1}{12}\text{, }&m=\frac{3}{2}\end{matrix}\right.$$