Solve (2x+1)^2-(x-5)(2x+1)=0

$$( 2 x + 1 ) ^ { 2 } - ( x - 5 ) ( 2 x + 1 ) = 0$$

Solve for x

$$x=-6$$

$$x=-\frac{1}{2}=-0.5$$

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}.

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

Use the distributive property to multiply x-5 by 2x+1 and combine like terms.

$$4x^{2}+4x+1-\left(2x^{2}-9x-5\right)=0$$

To find the opposite of 2x^{2}-9x-5, find the opposite of each term.

$$4x^{2}+4x+1-2x^{2}+9x+5=0$$

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

$$2x^{2}+4x+1+9x+5=0$$

Combine 4x and 9x to get 13x.

$$2x^{2}+13x+1+5=0$$

Add 1 and 5 to get 6.

$$2x^{2}+13x+6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 2x^{2}+ax+bx+6. To find a and b, set up a system to be solved.

$$a+b=13 ab=2\times 6=12$$

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 12.

$$1,12 2,6 3,4$$

Calculate the sum for each pair.

$$1+12=13 2+6=8 3+4=7$$

The solution is the pair that gives sum 13.

$$a=1 b=12$$

Rewrite 2x^{2}+13x+6 as \left(2x^{2}+x\right)+\left(12x+6\right).

$$\left(2x^{2}+x\right)+\left(12x+6\right)$$

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

$$x\left(2x+1\right)+6\left(2x+1\right)$$

Factor out common term 2x+1 by using distributive property.

$$\left(2x+1\right)\left(x+6\right)$$

To find equation solutions, solve 2x+1=0 and x+6=0.

$$x=-\frac{1}{2} x=-6$$

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}.

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

Use the distributive property to multiply x-5 by 2x+1 and combine like terms.

$$4x^{2}+4x+1-\left(2x^{2}-9x-5\right)=0$$

To find the opposite of 2x^{2}-9x-5, find the opposite of each term.

$$4x^{2}+4x+1-2x^{2}+9x+5=0$$

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

$$2x^{2}+4x+1+9x+5=0$$

Combine 4x and 9x to get 13x.

$$2x^{2}+13x+1+5=0$$

Add 1 and 5 to get 6.

$$2x^{2}+13x+6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 13 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-13±\sqrt{13^{2}-4\times 2\times 6}}{2\times 2}$$

Square 13.

$$x=\frac{-13±\sqrt{169-4\times 2\times 6}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-13±\sqrt{169-8\times 6}}{2\times 2}$$

Multiply -8 times 6.

$$x=\frac{-13±\sqrt{169-48}}{2\times 2}$$

Add 169 to -48.

$$x=\frac{-13±\sqrt{121}}{2\times 2}$$

Take the square root of 121.

$$x=\frac{-13±11}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-13±11}{4}$$

Now solve the equation x=\frac{-13±11}{4} when ± is plus. Add -13 to 11.

$$x=-\frac{2}{4}$$

Reduce the fraction \frac{-2}{4} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{2}$$

Now solve the equation x=\frac{-13±11}{4} when ± is minus. Subtract 11 from -13.

$$x=-\frac{24}{4}$$

Divide -24 by 4.

$$x=-6$$

The equation is now solved.

$$x=-\frac{1}{2} x=-6$$

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}.

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

Use the distributive property to multiply x-5 by 2x+1 and combine like terms.

$$4x^{2}+4x+1-\left(2x^{2}-9x-5\right)=0$$

To find the opposite of 2x^{2}-9x-5, find the opposite of each term.

$$4x^{2}+4x+1-2x^{2}+9x+5=0$$

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

$$2x^{2}+4x+1+9x+5=0$$

Combine 4x and 9x to get 13x.

$$2x^{2}+13x+1+5=0$$

Add 1 and 5 to get 6.

$$2x^{2}+13x+6=0$$

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

$$2x^{2}+13x=-6$$

Divide both sides by 2.

$$\frac{2x^{2}+13x}{2}=-\frac{6}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{13}{2}x=-\frac{6}{2}$$

Divide -6 by 2.

$$x^{2}+\frac{13}{2}x=-3$$

Divide \frac{13}{2}, the coefficient of the x term, by 2 to get \frac{13}{4}. Then add the square of \frac{13}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{13}{2}x+\left(\frac{13}{4}\right)^{2}=-3+\left(\frac{13}{4}\right)^{2}$$

Square \frac{13}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=-3+\frac{169}{16}$$

Add -3 to \frac{169}{16}.

$$x^{2}+\frac{13}{2}x+\frac{169}{16}=\frac{121}{16}$$

Factor x^{2}+\frac{13}{2}x+\frac{169}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{13}{4}\right)^{2}=\frac{121}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{13}{4}\right)^{2}}=\sqrt{\frac{121}{16}}$$

Simplify.

$$x+\frac{13}{4}=\frac{11}{4} x+\frac{13}{4}=-\frac{11}{4}$$

Subtract \frac{13}{4} from both sides of the equation.

$$x=-\frac{1}{2} x=-6$$