Solve (k-1)x^2+3kx+2k-3=0

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

Solve for k

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

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

Steps for Solving Linear Equation

Use the distributive property to multiply k-1 by x^{2}.

$$kx^{2}-x^{2}+3kx+2k-3=0$$

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

$$kx^{2}+3kx+2k-3=x^{2}$$

Add 3 to both sides.

$$kx^{2}+3kx+2k=x^{2}+3$$

Combine all terms containing k.

$$\left(x^{2}+3x+2\right)k=x^{2}+3$$

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

$$\frac{\left(x^{2}+3x+2\right)k}{x^{2}+3x+2}=\frac{x^{2}+3}{x^{2}+3x+2}$$

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

$$k=\frac{x^{2}+3}{x^{2}+3x+2}$$

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

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

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{k^{2}+20k-12}-3k}{2\left(k-1\right)}\text{; }x=-\frac{\sqrt{k^{2}+20k-12}+3k}{2\left(k-1\right)}\text{, }&k\neq 1\\x=\frac{1}{3}\text{, }&k=1\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{k^{2}+20k-12}-3k}{2\left(k-1\right)}\text{; }x=-\frac{\sqrt{k^{2}+20k-12}+3k}{2\left(k-1\right)}\text{, }&k\leq -4\sqrt{7}-10\text{ or }\left(k\neq 1\text{ and }k\geq 4\sqrt{7}-10\right)\\x=\frac{1}{3}\text{, }&k=1\end{matrix}\right.$$