Solve fx^2+4x+2=0

$$f x ^ { 2 } + 4 x + 2 = 0$$

Solve for f

$$f=-\frac{2\left(2x+1\right)}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Subtract 4x from both sides. Anything subtracted from zero gives its negation.

$$fx^{2}+2=-4x$$

Subtract 2 from both sides.

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

The equation is in standard form.

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

Divide both sides by x^{2}.

$$\frac{x^{2}f}{x^{2}}=\frac{-4x-2}{x^{2}}$$

Dividing by x^{2} undoes the multiplication by x^{2}.

$$f=\frac{-4x-2}{x^{2}}$$

Divide -4x-2 by x^{2}.

$$f=-\frac{2\left(2x+1\right)}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{2\left(2-f\right)}-2}{f}\text{; }x=-\frac{\sqrt{2}\left(\sqrt{2-f}+\sqrt{2}\right)}{f}\text{, }&f\neq 0\\x=-\frac{1}{2}\text{, }&f=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{2\left(2-f\right)}-2}{f}\text{; }x=-\frac{\sqrt{2}\left(\sqrt{2-f}+\sqrt{2}\right)}{f}\text{, }&f\neq 0\text{ and }f\leq 2\\x=-\frac{1}{2}\text{, }&f=0\end{matrix}\right.$$