Solve fc^2-12c+4=0

$$f c ^ { 2 } - 12 c + 4 = 0$$

Solve for f

$$f=-\frac{4\left(1-3c\right)}{c^{2}}$$

$$c\neq 0$$

Steps for Solving Linear Equation

Add 12c to both sides. Anything plus zero gives itself.

$$fc^{2}+4=12c$$

Subtract 4 from both sides.

$$fc^{2}=12c-4$$

The equation is in standard form.

$$c^{2}f=12c-4$$

Divide both sides by c^{2}.

$$\frac{c^{2}f}{c^{2}}=\frac{12c-4}{c^{2}}$$

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

$$f=\frac{12c-4}{c^{2}}$$

Divide 12c-4 by c^{2}.

$$f=\frac{4\left(3c-1\right)}{c^{2}}$$

Solve for c

$$\left\{\begin{matrix}c=\frac{2\left(\sqrt{9-f}+3\right)}{f}\text{; }c=\frac{2\left(-\sqrt{9-f}+3\right)}{f}\text{, }&f\neq 0\text{ and }f\leq 9\\c=\frac{1}{3}\approx 0.333333333\text{, }&f=0\end{matrix}\right.$$