Solve (2x+1)^2=(x-5)^2

$$( 2 x + 1 ) ^ { 2 } = ( x - 5 ) ^ { 2 }$$

Solve for x

$$x=-6$$

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

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}.

$$4x^{2}+4x+1=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$4x^{2}+4x+1=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$4x^{2}+4x+1-x^{2}=-10x+25$$

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

$$3x^{2}+4x+1=-10x+25$$

Add 10x to both sides.

$$3x^{2}+4x+1+10x=25$$

Combine 4x and 10x to get 14x.

$$3x^{2}+14x+1=25$$

Subtract 25 from both sides.

$$3x^{2}+14x+1-25=0$$

Subtract 25 from 1 to get -24.

$$3x^{2}+14x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=14 ab=3\left(-24\right)=-72$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -72.

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

Calculate the sum for each pair.

$$-1+72=71 -2+36=34 -3+24=21 -4+18=14 -6+12=6 -8+9=1$$

The solution is the pair that gives sum 14.

$$a=-4 b=18$$

Rewrite 3x^{2}+14x-24 as \left(3x^{2}-4x\right)+\left(18x-24\right).

$$\left(3x^{2}-4x\right)+\left(18x-24\right)$$

Factor out x in the first and 6 in the second group.

$$x\left(3x-4\right)+6\left(3x-4\right)$$

Factor out common term 3x-4 by using distributive property.

$$\left(3x-4\right)\left(x+6\right)$$

To find equation solutions, solve 3x-4=0 and x+6=0.

$$x=\frac{4}{3} x=-6$$

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}.

$$4x^{2}+4x+1=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$4x^{2}+4x+1=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$4x^{2}+4x+1-x^{2}=-10x+25$$

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

$$3x^{2}+4x+1=-10x+25$$

Add 10x to both sides.

$$3x^{2}+4x+1+10x=25$$

Combine 4x and 10x to get 14x.

$$3x^{2}+14x+1=25$$

Subtract 25 from both sides.

$$3x^{2}+14x+1-25=0$$

Subtract 25 from 1 to get -24.

$$3x^{2}+14x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 14 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-14±\sqrt{14^{2}-4\times 3\left(-24\right)}}{2\times 3}$$

Square 14.

$$x=\frac{-14±\sqrt{196-4\times 3\left(-24\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-14±\sqrt{196-12\left(-24\right)}}{2\times 3}$$

Multiply -12 times -24.

$$x=\frac{-14±\sqrt{196+288}}{2\times 3}$$

Add 196 to 288.

$$x=\frac{-14±\sqrt{484}}{2\times 3}$$

Take the square root of 484.

$$x=\frac{-14±22}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-14±22}{6}$$

Now solve the equation x=\frac{-14±22}{6} when ± is plus. Add -14 to 22.

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

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{-14±22}{6} when ± is minus. Subtract 22 from -14.

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

Divide -36 by 6.

$$x=-6$$

The equation is now solved.

$$x=\frac{4}{3} x=-6$$

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}.

$$4x^{2}+4x+1=\left(x-5\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-5\right)^{2}.

$$4x^{2}+4x+1=x^{2}-10x+25$$

Subtract x^{2} from both sides.

$$4x^{2}+4x+1-x^{2}=-10x+25$$

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

$$3x^{2}+4x+1=-10x+25$$

Add 10x to both sides.

$$3x^{2}+4x+1+10x=25$$

Combine 4x and 10x to get 14x.

$$3x^{2}+14x+1=25$$

Subtract 1 from both sides.

$$3x^{2}+14x=25-1$$

Subtract 1 from 25 to get 24.

$$3x^{2}+14x=24$$

Divide both sides by 3.

$$\frac{3x^{2}+14x}{3}=\frac{24}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{14}{3}x=\frac{24}{3}$$

Divide 24 by 3.

$$x^{2}+\frac{14}{3}x=8$$

Divide \frac{14}{3}, the coefficient of the x term, by 2 to get \frac{7}{3}. Then add the square of \frac{7}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{14}{3}x+\left(\frac{7}{3}\right)^{2}=8+\left(\frac{7}{3}\right)^{2}$$

Square \frac{7}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}=8+\frac{49}{9}$$

Add 8 to \frac{49}{9}.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}=\frac{121}{9}$$

Factor x^{2}+\frac{14}{3}x+\frac{49}{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}.

$$\left(x+\frac{7}{3}\right)^{2}=\frac{121}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{3}\right)^{2}}=\sqrt{\frac{121}{9}}$$

Simplify.

$$x+\frac{7}{3}=\frac{11}{3} x+\frac{7}{3}=-\frac{11}{3}$$

Subtract \frac{7}{3} from both sides of the equation.

$$x=\frac{4}{3} x=-6$$