Variable $k$ cannot be equal to $0$ since division by zero is not defined. Multiply both sides of the equation by $k$.
$$8=k\times 2+kk$$
Multiply $k$ and $k$ to get $k^{2}$.
$$8=k\times 2+k^{2}$$
Swap sides so that all variable terms are on the left hand side.
$$k\times 2+k^{2}=8$$
Subtract $8$ from both sides.
$$k\times 2+k^{2}-8=0$$
All equations of the form $ax^{2}+bx+c=0$ can be solved using the quadratic formula: $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$. The quadratic formula gives two solutions, one when $±$ is addition and one when it is subtraction.
$$k^{2}+2k-8=0$$
This equation is in standard form: $ax^{2}+bx+c=0$. Substitute $1$ for $a$, $2$ for $b$, and $-8$ for $c$ in the quadratic formula, $\frac{-b±\sqrt{b^{2}-4ac}}{2a}$.
$$k=\frac{-2±\sqrt{2^{2}-4\left(-8\right)}}{2}$$
Square $2$.
$$k=\frac{-2±\sqrt{4-4\left(-8\right)}}{2}$$
Multiply $-4$ times $-8$.
$$k=\frac{-2±\sqrt{4+32}}{2}$$
Add $4$ to $32$.
$$k=\frac{-2±\sqrt{36}}{2}$$
Take the square root of $36$.
$$k=\frac{-2±6}{2}$$
Now solve the equation $k=\frac{-2±6}{2}$ when $±$ is plus. Add $-2$ to $6$.
$$k=\frac{4}{2}$$
Divide $4$ by $2$.
$$k=2$$
Now solve the equation $k=\frac{-2±6}{2}$ when $±$ is minus. Subtract $6$ from $-2$.
$$k=-\frac{8}{2}$$
Divide $-8$ by $2$.
$$k=-4$$
The equation is now solved.
$$k=2$$ $$k=-4$$
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 $k$.
$$8=k\times 2+kk$$
Multiply $k$ and $k$ to get $k^{2}$.
$$8=k\times 2+k^{2}$$
Swap sides so that all variable terms are on the left hand side.
$$k\times 2+k^{2}=8$$
Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form $x^{2}+bx=c$.
$$k^{2}+2k=8$$
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.
$$k^{2}+2k+1^{2}=8+1^{2}$$
Square $1$.
$$k^{2}+2k+1=8+1$$
Add $8$ to $1$.
$$k^{2}+2k+1=9$$
Factor $k^{2}+2k+1$. In general, when $x^{2}+bx+c$ is a perfect square, it can always be factored as $\left(x+\frac{b}{2}\right)^{2}$.
$$\left(k+1\right)^{2}=9$$
Take the square root of both sides of the equation.
