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

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

Solve for k

$$k=-\frac{x^{2}-2x+3}{x-1}$$

$$x\neq 1$$

Steps for Solving Linear Equation

Use the distributive property to multiply k-2 by x.

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

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

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

Add 2x to both sides.

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

Subtract 3 from both sides.

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

Combine all terms containing k.

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

Divide both sides by x-1.

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

Dividing by x-1 undoes the multiplication by x-1.

$$k=\frac{-x^{2}+2x-3}{x-1}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{k^{2}-8}}{2}-\frac{k}{2}+1$$

$$x=-\frac{\sqrt{k^{2}-8}}{2}-\frac{k}{2}+1$$

Solve for x

$$x=\frac{\sqrt{k^{2}-8}}{2}-\frac{k}{2}+1$$

$$x=-\frac{\sqrt{k^{2}-8}}{2}-\frac{k}{2}+1\text{, }|k|\geq 2\sqrt{2}$$