Solve (2x+1)^2-(x+7)^2=0

$$( 2 x + 1 ) ^ { 2 } - ( x + 7 ) ^ { 2 } = 0$$

Solve for x

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

$$x=6$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

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

$$4x^{2}+4x+1-\left(x^{2}+14x+49\right)=0$$

To find the opposite of x^{2}+14x+49, find the opposite of each term.

$$4x^{2}+4x+1-x^{2}-14x-49=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+4x+1-14x-49=0$$

Combine 4x and -14x to get -10x.

$$3x^{2}-10x+1-49=0$$

Subtract 49 from 1 to get -48.

$$3x^{2}-10x-48=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-48. To find a and b, set up a system to be solved.

$$a+b=-10 ab=3\left(-48\right)=-144$$

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 -144.

$$1,-144 2,-72 3,-48 4,-36 6,-24 8,-18 9,-16 12,-12$$

Calculate the sum for each pair.

$$1-144=-143 2-72=-70 3-48=-45 4-36=-32 6-24=-18 8-18=-10 9-16=-7 12-12=0$$

The solution is the pair that gives sum -10.

$$a=-18 b=8$$

Rewrite 3x^{2}-10x-48 as \left(3x^{2}-18x\right)+\left(8x-48\right).

$$\left(3x^{2}-18x\right)+\left(8x-48\right)$$

Factor out 3x in the first and 8 in the second group.

$$3x\left(x-6\right)+8\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(3x+8\right)$$

To find equation solutions, solve x-6=0 and 3x+8=0.

$$x=6 x=-\frac{8}{3}$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

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

$$4x^{2}+4x+1-\left(x^{2}+14x+49\right)=0$$

To find the opposite of x^{2}+14x+49, find the opposite of each term.

$$4x^{2}+4x+1-x^{2}-14x-49=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+4x+1-14x-49=0$$

Combine 4x and -14x to get -10x.

$$3x^{2}-10x+1-49=0$$

Subtract 49 from 1 to get -48.

$$3x^{2}-10x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -10 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\times 3\left(-48\right)}}{2\times 3}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\times 3\left(-48\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-10\right)±\sqrt{100-12\left(-48\right)}}{2\times 3}$$

Multiply -12 times -48.

$$x=\frac{-\left(-10\right)±\sqrt{100+576}}{2\times 3}$$

Add 100 to 576.

$$x=\frac{-\left(-10\right)±\sqrt{676}}{2\times 3}$$

Take the square root of 676.

$$x=\frac{-\left(-10\right)±26}{2\times 3}$$

The opposite of -10 is 10.

$$x=\frac{10±26}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{10±26}{6}$$

Now solve the equation x=\frac{10±26}{6} when ± is plus. Add 10 to 26.

$$x=\frac{36}{6}$$

Divide 36 by 6.

$$x=6$$

Now solve the equation x=\frac{10±26}{6} when ± is minus. Subtract 26 from 10.

$$x=-\frac{16}{6}$$

Reduce the fraction \frac{-16}{6} to lowest terms by extracting and canceling out 2.

$$x=-\frac{8}{3}$$

The equation is now solved.

$$x=6 x=-\frac{8}{3}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(2x+1\right)^{2}.

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

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

$$4x^{2}+4x+1-\left(x^{2}+14x+49\right)=0$$

To find the opposite of x^{2}+14x+49, find the opposite of each term.

$$4x^{2}+4x+1-x^{2}-14x-49=0$$

Combine 4x^{2} and -x^{2} to get 3x^{2}.

$$3x^{2}+4x+1-14x-49=0$$

Combine 4x and -14x to get -10x.

$$3x^{2}-10x+1-49=0$$

Subtract 49 from 1 to get -48.

$$3x^{2}-10x-48=0$$

Add 48 to both sides. Anything plus zero gives itself.

$$3x^{2}-10x=48$$

Divide both sides by 3.

$$\frac{3x^{2}-10x}{3}=\frac{48}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{10}{3}x=\frac{48}{3}$$

Divide 48 by 3.

$$x^{2}-\frac{10}{3}x=16$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=16+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=16+\frac{25}{9}$$

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

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{169}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{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}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{169}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{169}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{13}{3} x-\frac{5}{3}=-\frac{13}{3}$$

Add \frac{5}{3} to both sides of the equation.

$$x=6 x=-\frac{8}{3}$$