Solve 0=k^2+12k-4

$$0 = k ^ { 2 } + 12 k - 4$$

Solve for k

$$k=2\sqrt{10}-6\approx 0.32455532$$

$$k=-2\sqrt{10}-6\approx -12.32455532$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$k^{2}+12k-4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 12 for b, and -4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-12±\sqrt{12^{2}-4\left(-4\right)}}{2}$$

Square 12.

$$k=\frac{-12±\sqrt{144-4\left(-4\right)}}{2}$$

Multiply -4 times -4.

$$k=\frac{-12±\sqrt{144+16}}{2}$$

Add 144 to 16.

$$k=\frac{-12±\sqrt{160}}{2}$$

Take the square root of 160.

$$k=\frac{-12±4\sqrt{10}}{2}$$

Now solve the equation k=\frac{-12±4\sqrt{10}}{2} when ± is plus. Add -12 to 4\sqrt{10}.

$$k=\frac{4\sqrt{10}-12}{2}$$

Divide -12+4\sqrt{10} by 2.

$$k=2\sqrt{10}-6$$

Now solve the equation k=\frac{-12±4\sqrt{10}}{2} when ± is minus. Subtract 4\sqrt{10} from -12.

$$k=\frac{-4\sqrt{10}-12}{2}$$

Divide -12-4\sqrt{10} by 2.

$$k=-2\sqrt{10}-6$$

The equation is now solved.

$$k=2\sqrt{10}-6 k=-2\sqrt{10}-6$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$k^{2}+12k-4=0$$

Add 4 to both sides. Anything plus zero gives itself.

$$k^{2}+12k=4$$

Divide 12, the coefficient of the x term, by 2 to get 6. Then add the square of 6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+12k+6^{2}=4+6^{2}$$

Square 6.

$$k^{2}+12k+36=4+36$$

Add 4 to 36.

$$k^{2}+12k+36=40$$

Factor k^{2}+12k+36. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+6\right)^{2}=40$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+6\right)^{2}}=\sqrt{40}$$

Simplify.

$$k+6=2\sqrt{10} k+6=-2\sqrt{10}$$

Subtract 6 from both sides of the equation.

$$k=2\sqrt{10}-6 k=-2\sqrt{10}-6$$