Solve 2x^2-kx+2=0

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

Solve for k

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

$$x\neq 0$$

Steps for Solving Linear Equation

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

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

Subtract 2 from both sides.

$$-kx=-2x^{2}-2$$

The equation is in standard form.

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

Divide both sides by -x.

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

Dividing by -x undoes the multiplication by -x.

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

Divide -2x^{2}-2 by -x.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{k^{2}-16}+k}{4}$$

$$x=\frac{-\sqrt{k^{2}-16}+k}{4}$$

Solve for x

$$x=\frac{\sqrt{k^{2}-16}+k}{4}$$

$$x=\frac{-\sqrt{k^{2}-16}+k}{4}\text{, }|k|\geq 4$$