Solve -k^2+k-6/2=0

$$- \frac { k ^ { 2 } + k - 6 } { 2 } = 0$$

Solve for k

$$k=-3$$

$$k=2$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 2.

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

To find the opposite of k^{2}+k-6, find the opposite of each term.

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -k^{2}+ak+bk+6. To find a and b, set up a system to be solved.

$$a+b=-1 ab=-6=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=2 b=-3$$

Rewrite -k^{2}-k+6 as \left(-k^{2}+2k\right)+\left(-3k+6\right).

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

Factor out k in the first and 3 in the second group.

$$k\left(-k+2\right)+3\left(-k+2\right)$$

Factor out common term -k+2 by using distributive property.

$$\left(-k+2\right)\left(k+3\right)$$

To find equation solutions, solve -k+2=0 and k+3=0.

$$k=2 k=-3$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 2.

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

To find the opposite of k^{2}+k-6, find the opposite of each term.

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

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

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

Multiply -4 times -1.

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

Multiply 4 times 6.

$$k=\frac{-\left(-1\right)±\sqrt{1+24}}{2\left(-1\right)}$$

Add 1 to 24.

$$k=\frac{-\left(-1\right)±\sqrt{25}}{2\left(-1\right)}$$

Take the square root of 25.

$$k=\frac{-\left(-1\right)±5}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$k=\frac{1±5}{2\left(-1\right)}$$

Multiply 2 times -1.

$$k=\frac{1±5}{-2}$$

Now solve the equation k=\frac{1±5}{-2} when ± is plus. Add 1 to 5.

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

Divide 6 by -2.

$$k=-3$$

Now solve the equation k=\frac{1±5}{-2} when ± is minus. Subtract 5 from 1.

$$k=-\frac{4}{-2}$$

Divide -4 by -2.

$$k=2$$

The equation is now solved.

$$k=-3 k=2$$

Steps for Completing the Square

Multiply both sides of the equation by 2.

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

To find the opposite of k^{2}+k-6, find the opposite of each term.

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

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide -6 by -1.

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

Divide 1, the coefficient of the x term, by 2 to get \frac{1}{2}. Then add the square of \frac{1}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$k^{2}+k+\frac{1}{4}=6+\frac{1}{4}$$

Add 6 to \frac{1}{4}.

$$k^{2}+k+\frac{1}{4}=\frac{25}{4}$$

Factor k^{2}+k+\frac{1}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+\frac{1}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+\frac{1}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$k+\frac{1}{2}=\frac{5}{2} k+\frac{1}{2}=-\frac{5}{2}$$

Subtract \frac{1}{2} from both sides of the equation.

$$k=2 k=-3$$