Solve -20/k-6=-k/2

$$\frac { - 20 } { k - 6 } = \frac { - k } { 2 }$$

Solve for k

$$k=-4$$

$$k=10$$

Steps Using the Quadratic Formula

Variable k cannot be equal to 6 since division by zero is not defined. Multiply both sides of the equation by 2\left(k-6\right), the least common multiple of k-6,2.

$$2\left(-20\right)=\left(k-6\right)\left(-k\right)$$

Multiply 2 and -20 to get -40.

$$-40=\left(k-6\right)\left(-k\right)$$

Use the distributive property to multiply k-6 by -k.

$$-40=k\left(-k\right)-6\left(-k\right)$$

Multiply -6 and -1 to get 6.

$$-40=k\left(-k\right)+6k$$

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

$$k\left(-k\right)+6k=-40$$

Add 40 to both sides.

$$k\left(-k\right)+6k+40=0$$

Multiply k and k to get k^{2}.

$$k^{2}\left(-1\right)+6k+40=0$$

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.

$$-k^{2}+6k+40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 6 for b, and 40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-6±\sqrt{6^{2}-4\left(-1\right)\times 40}}{2\left(-1\right)}$$

Square 6.

$$k=\frac{-6±\sqrt{36-4\left(-1\right)\times 40}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$k=\frac{-6±\sqrt{36+4\times 40}}{2\left(-1\right)}$$

Multiply 4 times 40.

$$k=\frac{-6±\sqrt{36+160}}{2\left(-1\right)}$$

Add 36 to 160.

$$k=\frac{-6±\sqrt{196}}{2\left(-1\right)}$$

Take the square root of 196.

$$k=\frac{-6±14}{2\left(-1\right)}$$

Multiply 2 times -1.

$$k=\frac{-6±14}{-2}$$

Now solve the equation k=\frac{-6±14}{-2} when ± is plus. Add -6 to 14.

$$k=\frac{8}{-2}$$

Divide 8 by -2.

$$k=-4$$

Now solve the equation k=\frac{-6±14}{-2} when ± is minus. Subtract 14 from -6.

$$k=-\frac{20}{-2}$$

Divide -20 by -2.

$$k=10$$

The equation is now solved.

$$k=-4 k=10$$

Steps for Completing the Square

Variable k cannot be equal to 6 since division by zero is not defined. Multiply both sides of the equation by 2\left(k-6\right), the least common multiple of k-6,2.

$$2\left(-20\right)=\left(k-6\right)\left(-k\right)$$

Multiply 2 and -20 to get -40.

$$-40=\left(k-6\right)\left(-k\right)$$

Use the distributive property to multiply k-6 by -k.

$$-40=k\left(-k\right)-6\left(-k\right)$$

Multiply -6 and -1 to get 6.

$$-40=k\left(-k\right)+6k$$

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

$$k\left(-k\right)+6k=-40$$

Multiply k and k to get k^{2}.

$$k^{2}\left(-1\right)+6k=-40$$

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.

$$-k^{2}+6k=-40$$

Divide both sides by -1.

$$\frac{-k^{2}+6k}{-1}=-\frac{40}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$k^{2}+\frac{6}{-1}k=-\frac{40}{-1}$$

Divide 6 by -1.

$$k^{2}-6k=-\frac{40}{-1}$$

Divide -40 by -1.

$$k^{2}-6k=40$$

Divide -6, the coefficient of the x term, by 2 to get -3. Then add the square of -3 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}-6k+\left(-3\right)^{2}=40+\left(-3\right)^{2}$$

Square -3.

$$k^{2}-6k+9=40+9$$

Add 40 to 9.

$$k^{2}-6k+9=49$$

Factor k^{2}-6k+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k-3\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k-3\right)^{2}}=\sqrt{49}$$

Simplify.

$$k-3=7 k-3=-7$$

Add 3 to both sides of the equation.

$$k=10 k=-4$$