Solve 5/15kx^2+4x-2=0

$$\frac { 5 } { 15 } k x ^ { 2 } + 4 x - 2 = 0$$

Solve for k

$$k=-\frac{6\left(2x-1\right)}{x^{2}}$$

$$x\neq 0$$

Steps for Solving Linear Equation

Reduce the fraction \frac{5}{15} to lowest terms by extracting and canceling out 5.

$$\frac{1}{3}kx^{2}+4x-2=0$$

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

$$\frac{1}{3}kx^{2}-2=-4x$$

Add 2 to both sides.

$$\frac{1}{3}kx^{2}=-4x+2$$

The equation is in standard form.

$$\frac{x^{2}}{3}k=2-4x$$

Divide both sides by \frac{1}{3}x^{2}.

$$\frac{3\times \frac{x^{2}}{3}k}{x^{2}}=\frac{3\left(2-4x\right)}{x^{2}}$$

Dividing by \frac{1}{3}x^{2} undoes the multiplication by \frac{1}{3}x^{2}.

$$k=\frac{3\left(2-4x\right)}{x^{2}}$$

Divide -4x+2 by \frac{1}{3}x^{2}.

$$k=\frac{6\left(1-2x\right)}{x^{2}}$$

Solve for x (complex solution)

$$\left\{\begin{matrix}x=\frac{\sqrt{6\left(k+6\right)}-6}{k}\text{; }x=-\frac{\sqrt{6}\left(\sqrt{k+6}+\sqrt{6}\right)}{k}\text{, }&k\neq 0\\x=\frac{1}{2}\text{, }&k=0\end{matrix}\right.$$

Solve for x

$$\left\{\begin{matrix}x=\frac{\sqrt{6\left(k+6\right)}-6}{k}\text{; }x=-\frac{\sqrt{6}\left(\sqrt{k+6}+\sqrt{6}\right)}{k}\text{, }&k\neq 0\text{ and }k\geq -6\\x=\frac{1}{2}\text{, }&k=0\end{matrix}\right.$$