Solve (-12-k)^2-4(4)(4)=0

$$( - 12 - k ) ^ { 2 } - 4 ( 4 ) ( 4 ) = 0$$

Solve for k

$$k=-20$$

$$k=-4$$

Steps Using Factoring

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-12-k\right)^{2}.

$$144+24k+k^{2}-4\times 4\times 4=0$$

Multiply 4 and 4 to get 16.

$$144+24k+k^{2}-16\times 4=0$$

Multiply 16 and 4 to get 64.

$$144+24k+k^{2}-64=0$$

Subtract 64 from 144 to get 80.

$$80+24k+k^{2}=0$$

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

$$k^{2}+24k+80=0$$

To solve the equation, factor k^{2}+24k+80 using formula k^{2}+\left(a+b\right)k+ab=\left(k+a\right)\left(k+b\right). To find a and b, set up a system to be solved.

$$a+b=24 ab=80$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 80.

$$1,80 2,40 4,20 5,16 8,10$$

Calculate the sum for each pair.

$$1+80=81 2+40=42 4+20=24 5+16=21 8+10=18$$

The solution is the pair that gives sum 24.

$$a=4 b=20$$

Rewrite factored expression \left(k+a\right)\left(k+b\right) using the obtained values.

$$\left(k+4\right)\left(k+20\right)$$

To find equation solutions, solve k+4=0 and k+20=0.

$$k=-4 k=-20$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-12-k\right)^{2}.

$$144+24k+k^{2}-4\times 4\times 4=0$$

Multiply 4 and 4 to get 16.

$$144+24k+k^{2}-16\times 4=0$$

Multiply 16 and 4 to get 64.

$$144+24k+k^{2}-64=0$$

Subtract 64 from 144 to get 80.

$$80+24k+k^{2}=0$$

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

$$k^{2}+24k+80=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as k^{2}+ak+bk+80. To find a and b, set up a system to be solved.

$$a+b=24 ab=1\times 80=80$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 80.

$$1,80 2,40 4,20 5,16 8,10$$

Calculate the sum for each pair.

$$1+80=81 2+40=42 4+20=24 5+16=21 8+10=18$$

The solution is the pair that gives sum 24.

$$a=4 b=20$$

Rewrite k^{2}+24k+80 as \left(k^{2}+4k\right)+\left(20k+80\right).

$$\left(k^{2}+4k\right)+\left(20k+80\right)$$

Factor out k in the first and 20 in the second group.

$$k\left(k+4\right)+20\left(k+4\right)$$

Factor out common term k+4 by using distributive property.

$$\left(k+4\right)\left(k+20\right)$$

To find equation solutions, solve k+4=0 and k+20=0.

$$k=-4 k=-20$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-12-k\right)^{2}.

$$144+24k+k^{2}-4\times 4\times 4=0$$

Multiply 4 and 4 to get 16.

$$144+24k+k^{2}-16\times 4=0$$

Multiply 16 and 4 to get 64.

$$144+24k+k^{2}-64=0$$

Subtract 64 from 144 to get 80.

$$80+24k+k^{2}=0$$

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.

$$k^{2}+24k+80=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, 24 for b, and 80 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-24±\sqrt{24^{2}-4\times 80}}{2}$$

Square 24.

$$k=\frac{-24±\sqrt{576-4\times 80}}{2}$$

Multiply -4 times 80.

$$k=\frac{-24±\sqrt{576-320}}{2}$$

Add 576 to -320.

$$k=\frac{-24±\sqrt{256}}{2}$$

Take the square root of 256.

$$k=\frac{-24±16}{2}$$

Now solve the equation k=\frac{-24±16}{2} when ± is plus. Add -24 to 16.

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

Divide -8 by 2.

$$k=-4$$

Now solve the equation k=\frac{-24±16}{2} when ± is minus. Subtract 16 from -24.

$$k=-\frac{40}{2}$$

Divide -40 by 2.

$$k=-20$$

The equation is now solved.

$$k=-4 k=-20$$

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(-12-k\right)^{2}.

$$144+24k+k^{2}-4\times 4\times 4=0$$

Multiply 4 and 4 to get 16.

$$144+24k+k^{2}-16\times 4=0$$

Multiply 16 and 4 to get 64.

$$144+24k+k^{2}-64=0$$

Subtract 64 from 144 to get 80.

$$80+24k+k^{2}=0$$

Subtract 80 from both sides. Anything subtracted from zero gives its negation.

$$24k+k^{2}=-80$$

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.

$$k^{2}+24k=-80$$

Divide 24, the coefficient of the x term, by 2 to get 12. Then add the square of 12 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$k^{2}+24k+12^{2}=-80+12^{2}$$

Square 12.

$$k^{2}+24k+144=-80+144$$

Add -80 to 144.

$$k^{2}+24k+144=64$$

Factor k^{2}+24k+144. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(k+12\right)^{2}=64$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+12\right)^{2}}=\sqrt{64}$$

Simplify.

$$k+12=8 k+12=-8$$

Subtract 12 from both sides of the equation.

$$k=-4 k=-20$$