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

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

Solve for k (complex solution)

$$\left\{\begin{matrix}k=-\frac{1-x}{x+1}\text{, }&x\neq -1\\k\in \mathrm{C}\text{, }&x=-2\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

Use the distributive property to multiply 3k-1 by x.

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

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

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

Add x to both sides.

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

Subtract 2 from both sides.

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

Combine all terms containing k.

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

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

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

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

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

Divide \left(-1+x\right)\left(2+x\right) by x^{2}+3x+2.

$$k=\frac{x-1}{x+1}$$

Solve for k

$$\left\{\begin{matrix}k=-\frac{1-x}{x+1}\text{, }&x\neq -1\\k\in \mathrm{R}\text{, }&x=-2\end{matrix}\right.$$

Steps for Solving Linear Equation

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

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

Use the distributive property to multiply 3k-1 by x.

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

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

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

Add x to both sides.

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

Subtract 2 from both sides.

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

Combine all terms containing k.

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

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

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

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

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

Divide \left(-1+x\right)\left(2+x\right) by x^{2}+3x+2.

$$k=\frac{x-1}{x+1}$$

Solve for x

$$\left\{\begin{matrix}\\x=-2\text{, }&\text{unconditionally}\\x=-\frac{k+1}{k-1}\text{, }&k\neq 1\end{matrix}\right.$$