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

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

Solve for k

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

$$x\neq 1$$

Steps for Solving Linear Equation

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

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

To find the opposite of kx-3x, find the opposite of each term.

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

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

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

Subtract 3x from both sides.

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

Add 1 to both sides.

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

Combine all terms containing k.

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

The equation is in standard form.

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

Divide both sides by -x+1.

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

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

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

Solve for x (complex solution)

$$x=\frac{\sqrt{k^{2}-14k+17}+k-3}{4}$$

$$x=\frac{-\sqrt{k^{2}-14k+17}+k-3}{4}$$

Solve for x

$$x=\frac{\sqrt{k^{2}-14k+17}+k-3}{4}$$

$$x=\frac{-\sqrt{k^{2}-14k+17}+k-3}{4}\text{, }k\geq 4\sqrt{2}+7\text{ or }k\leq 7-4\sqrt{2}$$