Solve 3x^2+kx+8=0

$$3 x ^ { 2 } + k x + 8 = 0$$

Solve for k

$$k=-3x-\frac{8}{x}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 3x^{2} from both sides. Anything subtracted from zero gives its negation.

$$kx+8=-3x^{2}$$

Subtract 8 from both sides.

$$kx=-3x^{2}-8$$

The equation is in standard form.

$$xk=-3x^{2}-8$$

Divide both sides by x.

$$\frac{xk}{x}=\frac{-3x^{2}-8}{x}$$

Dividing by x undoes the multiplication by x.

$$k=\frac{-3x^{2}-8}{x}$$

Divide -3x^{2}-8 by x.

$$k=-3x-\frac{8}{x}$$

Solve for x (complex solution)

$$x=\frac{\sqrt{k^{2}-96}-k}{6}$$

$$x=\frac{-\sqrt{k^{2}-96}-k}{6}$$

Solve for x

$$x=\frac{\sqrt{k^{2}-96}-k}{6}$$

$$x=\frac{-\sqrt{k^{2}-96}-k}{6}\text{, }|k|\geq 4\sqrt{6}$$