Variable $k$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $k$.
$$k^{2}+k^{2}+k^{2}=6k$$
Combine $k^{2}$ and $k^{2}$ to get $2k^{2}$.
$$2k^{2}+k^{2}=6k$$
Combine $2k^{2}$ and $k^{2}$ to get $3k^{2}$.
$$3k^{2}=6k$$
Subtract $6k$ from both sides.
$$3k^{2}-6k=0$$
Factor out $k$.
$$k\left(3k-6\right)=0$$
To find equation solutions, solve $k=0$ and $3k-6=0$.
$$k=0$$ $$k=2$$
Variable $k$ cannot be equal to $0$.
$$k=2$$
Steps Using the Quadratic Formula
Variable $k$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $k$.
$$k^{2}+k^{2}+k^{2}=6k$$
Combine $k^{2}$ and $k^{2}$ to get $2k^{2}$.
$$2k^{2}+k^{2}=6k$$
Combine $2k^{2}$ and $k^{2}$ to get $3k^{2}$.
$$3k^{2}=6k$$
Subtract $6k$ from both sides.
$$3k^{2}-6k=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $3$ for $a$, $-6$ for $b$, and $0$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
Now solve the equation $k=\frac{6±6}{6}$ when $±$ is plus. Add $6$ to $6$.
$$k=\frac{12}{6}$$
Divide $12$ by $6$.
$$k=2$$
Now solve the equation $k=\frac{6±6}{6}$ when $±$ is minus. Subtract $6$ from $6$.
$$k=\frac{0}{6}$$
Divide $0$ by $6$.
$$k=0$$
The equation is now solved.
$$k=2$$ $$k=0$$
Variable $k$ cannot be equal to $0$.
$$k=2$$
Steps for Completing the Square
Variable $k$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $k$.
$$k^{2}+k^{2}+k^{2}=6k$$
Combine $k^{2}$ and $k^{2}$ to get $2k^{2}$.
$$2k^{2}+k^{2}=6k$$
Combine $2k^{2}$ and $k^{2}$ to get $3k^{2}$.
$$3k^{2}=6k$$
Subtract $6k$ from both sides.
$$3k^{2}-6k=0$$
Divide both sides by $3$.
$$\frac{3k^{2}-6k}{3}=\frac{0}{3}$$
Dividing by $3$ undoes the multiplication by $3$.
$$k^{2}+\left(-\frac{6}{3}\right)k=\frac{0}{3}$$
Divide $-6$ by $3$.
$$k^{2}-2k=\frac{0}{3}$$
Divide $0$ by $3$.
$$k^{2}-2k=0$$
Divide $-2$, the coefficient of the $x$ term, by $2$ to get $-1$. Then add the square of $-1$ to both sides of the equation. This step makes the left hand side of the equation a perfect square.
$$k^{2}-2k+1=1$$
Factor $k^{2}-2k+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(k-1\right)^{2}=1$$
Take the square root of both sides of the equation.
