Solve (2x-1)^2=(3x+1)^2

$$( 2 x - 1 ) ^ { 2 } = ( 3 x + 1 ) ^ { 2 }$$

Solve for x

$$x=-2$$

$$x=0$$

Steps Using Factoring

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

$$4x^{2}-4x+1=\left(3x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$4x^{2}-4x+1=9x^{2}+6x+1$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+1-9x^{2}=6x+1$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-4x+1=6x+1$$

Subtract 6x from both sides.

$$-5x^{2}-4x+1-6x=1$$

Combine -4x and -6x to get -10x.

$$-5x^{2}-10x+1=1$$

Subtract 1 from both sides.

$$-5x^{2}-10x+1-1=0$$

Subtract 1 from 1 to get 0.

$$-5x^{2}-10x=0$$

Factor out x.

$$x\left(-5x-10\right)=0$$

To find equation solutions, solve x=0 and -5x-10=0.

$$x=0 x=-2$$

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}.

$$4x^{2}-4x+1=\left(3x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$4x^{2}-4x+1=9x^{2}+6x+1$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+1-9x^{2}=6x+1$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-4x+1=6x+1$$

Subtract 6x from both sides.

$$-5x^{2}-4x+1-6x=1$$

Combine -4x and -6x to get -10x.

$$-5x^{2}-10x+1=1$$

Subtract 1 from both sides.

$$-5x^{2}-10x+1-1=0$$

Subtract 1 from 1 to get 0.

$$-5x^{2}-10x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -10 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}}}{2\left(-5\right)}$$

Take the square root of \left(-10\right)^{2}.

$$x=\frac{-\left(-10\right)±10}{2\left(-5\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±10}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{10±10}{-10}$$

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

$$x=\frac{20}{-10}$$

Divide 20 by -10.

$$x=-2$$

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

$$x=\frac{0}{-10}$$

Divide 0 by -10.

$$x=0$$

The equation is now solved.

$$x=-2 x=0$$

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}.

$$4x^{2}-4x+1=\left(3x+1\right)^{2}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$4x^{2}-4x+1=9x^{2}+6x+1$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+1-9x^{2}=6x+1$$

Combine 4x^{2} and -9x^{2} to get -5x^{2}.

$$-5x^{2}-4x+1=6x+1$$

Subtract 6x from both sides.

$$-5x^{2}-4x+1-6x=1$$

Combine -4x and -6x to get -10x.

$$-5x^{2}-10x+1=1$$

Subtract 1 from both sides.

$$-5x^{2}-10x=1-1$$

Subtract 1 from 1 to get 0.

$$-5x^{2}-10x=0$$

Divide both sides by -5.

$$\frac{-5x^{2}-10x}{-5}=\frac{0}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\left(-\frac{10}{-5}\right)x=\frac{0}{-5}$$

Divide -10 by -5.

$$x^{2}+2x=\frac{0}{-5}$$

Divide 0 by -5.

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

Divide 2, the coefficient of the x term, by 2 to get 1. Then add the square of 1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+2x+1^{2}=1^{2}$$

Square 1.

$$x^{2}+2x+1=1$$

Factor x^{2}+2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x+1\right)^{2}}=\sqrt{1}$$

Simplify.

$$x+1=1 x+1=-1$$

Subtract 1 from both sides of the equation.

$$x=0 x=-2$$