Solve 2x^2-8x-2*K=0

$$2 x ^ { 2 } - 8 x - 2 \cdot K = 0$$

Solve for K

$$K=x\left(x-4\right)$$

Steps for Solving Linear Equation

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

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

Add 8x to both sides.

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

The equation is in standard form.

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

Divide both sides by -2.

$$\frac{-2K}{-2}=\frac{2x\left(4-x\right)}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$K=\frac{2x\left(4-x\right)}{-2}$$

Divide 2x\left(4-x\right) by -2.

$$K=x^{2}-4x$$

Solve for x (complex solution)

$$x=\sqrt{K+4}+2$$

$$x=-\sqrt{K+4}+2$$

Solve for x

$$x=\sqrt{K+4}+2$$

$$x=-\sqrt{K+4}+2\text{, }K\geq -4$$