Solve 2x^2-kx+x+8=0

$$2 x ^ { 2 } - k x + x + 8 = 0$$

Solve for k

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

$$x\neq 0$$

Steps for Solving Linear Equation

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

$$-kx+x+8=-2x^{2}$$

Subtract x from both sides.

$$-kx+8=-2x^{2}-x$$

Subtract 8 from both sides.

$$-kx=-2x^{2}-x-8$$

The equation is in standard form.

$$\left(-x\right)k=-2x^{2}-x-8$$

Divide both sides by -x.

$$\frac{\left(-x\right)k}{-x}=\frac{-2x^{2}-x-8}{-x}$$

Dividing by -x undoes the multiplication by -x.

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

Divide -2x^{2}-x-8 by -x.

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

Solve for x (complex solution)

$$x=\frac{\sqrt{\left(k-9\right)\left(k+7\right)}+k-1}{4}$$

$$x=\frac{-\sqrt{\left(k-9\right)\left(k+7\right)}+k-1}{4}$$

Solve for x

$$x=\frac{\sqrt{\left(k-9\right)\left(k+7\right)}+k-1}{4}$$

$$x=\frac{-\sqrt{\left(k-9\right)\left(k+7\right)}+k-1}{4}\text{, }k\leq -7\text{ or }k\geq 9$$