Solve left(7+2kright)^2-4left(6+kright)left(5k+1right)=0

$${ \left(7+2k \right) }^{ 2 } -4 \left( 6+k \right) \left( 5k+1 \right) =0$$

Solve for k

$$k = -\frac{25}{4} = -6\frac{1}{4} = -6.25$$

$$k=\frac{1}{4}=0.25$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+2k\right)^{2}.

$$49+28k+4k^{2}-4\left(6+k\right)\left(5k+1\right)=0$$

Use the distributive property to multiply -4 by 6+k.

$$49+28k+4k^{2}+\left(-24-4k\right)\left(5k+1\right)=0$$

Use the distributive property to multiply -24-4k by 5k+1 and combine like terms.

$$49+28k+4k^{2}-124k-24-20k^{2}=0$$

Combine 28k and -124k to get -96k.

$$49-96k+4k^{2}-24-20k^{2}=0$$

Subtract 24 from 49 to get 25.

$$25-96k+4k^{2}-20k^{2}=0$$

Combine 4k^{2} and -20k^{2} to get -16k^{2}.

$$25-96k-16k^{2}=0$$

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

$$-16k^{2}-96k+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -16k^{2}+ak+bk+25. To find a and b, set up a system to be solved.

$$a+b=-96 ab=-16\times 25=-400$$

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 -400.

$$1,-400 2,-200 4,-100 5,-80 8,-50 10,-40 16,-25 20,-20$$

Calculate the sum for each pair.

$$1-400=-399 2-200=-198 4-100=-96 5-80=-75 8-50=-42 10-40=-30 16-25=-9 20-20=0$$

The solution is the pair that gives sum -96.

$$a=4 b=-100$$

Rewrite -16k^{2}-96k+25 as \left(-16k^{2}+4k\right)+\left(-100k+25\right).

$$\left(-16k^{2}+4k\right)+\left(-100k+25\right)$$

Factor out 4k in the first and 25 in the second group.

$$4k\left(-4k+1\right)+25\left(-4k+1\right)$$

Factor out common term -4k+1 by using distributive property.

$$\left(-4k+1\right)\left(4k+25\right)$$

To find equation solutions, solve -4k+1=0 and 4k+25=0.

$$k=\frac{1}{4} k=-\frac{25}{4}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+2k\right)^{2}.

$$49+28k+4k^{2}-4\left(6+k\right)\left(5k+1\right)=0$$

Use the distributive property to multiply -4 by 6+k.

$$49+28k+4k^{2}+\left(-24-4k\right)\left(5k+1\right)=0$$

Use the distributive property to multiply -24-4k by 5k+1 and combine like terms.

$$49+28k+4k^{2}-124k-24-20k^{2}=0$$

Combine 28k and -124k to get -96k.

$$49-96k+4k^{2}-24-20k^{2}=0$$

Subtract 24 from 49 to get 25.

$$25-96k+4k^{2}-20k^{2}=0$$

Combine 4k^{2} and -20k^{2} to get -16k^{2}.

$$25-96k-16k^{2}=0$$

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.

$$-16k^{2}-96k+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -16 for a, -96 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$k=\frac{-\left(-96\right)±\sqrt{\left(-96\right)^{2}-4\left(-16\right)\times 25}}{2\left(-16\right)}$$

Square -96.

$$k=\frac{-\left(-96\right)±\sqrt{9216-4\left(-16\right)\times 25}}{2\left(-16\right)}$$

Multiply -4 times -16.

$$k=\frac{-\left(-96\right)±\sqrt{9216+64\times 25}}{2\left(-16\right)}$$

Multiply 64 times 25.

$$k=\frac{-\left(-96\right)±\sqrt{9216+1600}}{2\left(-16\right)}$$

Add 9216 to 1600.

$$k=\frac{-\left(-96\right)±\sqrt{10816}}{2\left(-16\right)}$$

Take the square root of 10816.

$$k=\frac{-\left(-96\right)±104}{2\left(-16\right)}$$

The opposite of -96 is 96.

$$k=\frac{96±104}{2\left(-16\right)}$$

Multiply 2 times -16.

$$k=\frac{96±104}{-32}$$

Now solve the equation k=\frac{96±104}{-32} when ± is plus. Add 96 to 104.

$$k=\frac{200}{-32}$$

Reduce the fraction \frac{200}{-32} to lowest terms by extracting and canceling out 8.

$$k=-\frac{25}{4}$$

Now solve the equation k=\frac{96±104}{-32} when ± is minus. Subtract 104 from 96.

$$k=-\frac{8}{-32}$$

Reduce the fraction \frac{-8}{-32} to lowest terms by extracting and canceling out 8.

$$k=\frac{1}{4}$$

The equation is now solved.

$$k=-\frac{25}{4} k=\frac{1}{4}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(7+2k\right)^{2}.

$$49+28k+4k^{2}-4\left(6+k\right)\left(5k+1\right)=0$$

Use the distributive property to multiply -4 by 6+k.

$$49+28k+4k^{2}+\left(-24-4k\right)\left(5k+1\right)=0$$

Use the distributive property to multiply -24-4k by 5k+1 and combine like terms.

$$49+28k+4k^{2}-124k-24-20k^{2}=0$$

Combine 28k and -124k to get -96k.

$$49-96k+4k^{2}-24-20k^{2}=0$$

Subtract 24 from 49 to get 25.

$$25-96k+4k^{2}-20k^{2}=0$$

Combine 4k^{2} and -20k^{2} to get -16k^{2}.

$$25-96k-16k^{2}=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$-96k-16k^{2}=-25$$

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.

$$-16k^{2}-96k=-25$$

Divide both sides by -16.

$$\frac{-16k^{2}-96k}{-16}=-\frac{25}{-16}$$

Dividing by -16 undoes the multiplication by -16.

$$k^{2}+\left(-\frac{96}{-16}\right)k=-\frac{25}{-16}$$

Divide -96 by -16.

$$k^{2}+6k=-\frac{25}{-16}$$

Divide -25 by -16.

$$k^{2}+6k=\frac{25}{16}$$

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.

$$k^{2}+6k+3^{2}=\frac{25}{16}+3^{2}$$

Square 3.

$$k^{2}+6k+9=\frac{25}{16}+9$$

Add \frac{25}{16} to 9.

$$k^{2}+6k+9=\frac{169}{16}$$

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}.

$$\left(k+3\right)^{2}=\frac{169}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(k+3\right)^{2}}=\sqrt{\frac{169}{16}}$$

Simplify.

$$k+3=\frac{13}{4} k+3=-\frac{13}{4}$$

Subtract 3 from both sides of the equation.

$$k=\frac{1}{4} k=-\frac{25}{4}$$