Solve 4-7=-2k-k^2

$$4 - 7 = - 2 k - k ^ { 2 }$$

Solve for k

$$k=-3$$

$$k=1$$

Steps Using Factoring By Grouping

Subtract 7 from 4 to get -3.

$$-3=-2k-k^{2}$$

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

$$-2k-k^{2}=-3$$

Add 3 to both sides.

$$-2k-k^{2}+3=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-k^{2}-2k+3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -k^{2}+ak+bk+3. To find a and b, set up a system to be solved.

$$a+b=-2 ab=-3=-3$$

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. The only such pair is the system solution.

$$a=1 b=-3$$

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

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

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

$$k\left(-k+1\right)+3\left(-k+1\right)$$

Factor out common term -k+1 by using distributive property.

$$\left(-k+1\right)\left(k+3\right)$$

To find equation solutions, solve -k+1=0 and k+3=0.

$$k=1 k=-3$$

Steps Using the Quadratic Formula

Subtract 7 from 4 to get -3.

$$-3=-2k-k^{2}$$

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

$$-2k-k^{2}=-3$$

Add 3 to both sides.

$$-2k-k^{2}+3=0$$

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.

$$-k^{2}-2k+3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, -2 for b, and 3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Square -2.

$$k=\frac{-\left(-2\right)±\sqrt{4-4\left(-1\right)\times 3}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$k=\frac{-\left(-2\right)±\sqrt{4+4\times 3}}{2\left(-1\right)}$$

Multiply 4 times 3.

$$k=\frac{-\left(-2\right)±\sqrt{4+12}}{2\left(-1\right)}$$

Add 4 to 12.

$$k=\frac{-\left(-2\right)±\sqrt{16}}{2\left(-1\right)}$$

Take the square root of 16.

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

The opposite of -2 is 2.

$$k=\frac{2±4}{2\left(-1\right)}$$

Multiply 2 times -1.

$$k=\frac{2±4}{-2}$$

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

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

Divide 6 by -2.

$$k=-3$$

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

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

Divide -2 by -2.

$$k=1$$

The equation is now solved.

$$k=-3 k=1$$

Steps for Completing the Square

Subtract 7 from 4 to get -3.

$$-3=-2k-k^{2}$$

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

$$-2k-k^{2}=-3$$

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.

$$-k^{2}-2k=-3$$

Divide both sides by -1.

$$\frac{-k^{2}-2k}{-1}=-\frac{3}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$k^{2}+\left(-\frac{2}{-1}\right)k=-\frac{3}{-1}$$

Divide -2 by -1.

$$k^{2}+2k=-\frac{3}{-1}$$

Divide -3 by -1.

$$k^{2}+2k=3$$

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.

$$k^{2}+2k+1^{2}=3+1^{2}$$

Square 1.

$$k^{2}+2k+1=3+1$$

Add 3 to 1.

$$k^{2}+2k+1=4$$

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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$k+1=2 k+1=-2$$

Subtract 1 from both sides of the equation.

$$k=1 k=-3$$