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

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

Solve for x

$$x=-\frac{1}{5}=-0.2$$

$$x=-3$$

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}.

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

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

$$4x^{2}-4x+1=9x^{2}+12x+4$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+1-9x^{2}=12x+4$$

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

$$-5x^{2}-4x+1=12x+4$$

Subtract 12x from both sides.

$$-5x^{2}-4x+1-12x=4$$

Combine -4x and -12x to get -16x.

$$-5x^{2}-16x+1=4$$

Subtract 4 from both sides.

$$-5x^{2}-16x+1-4=0$$

Subtract 4 from 1 to get -3.

$$-5x^{2}-16x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -5x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-16 ab=-5\left(-3\right)=15$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 15.

$$-1,-15 -3,-5$$

Calculate the sum for each pair.

$$-1-15=-16 -3-5=-8$$

The solution is the pair that gives sum -16.

$$a=-1 b=-15$$

Rewrite -5x^{2}-16x-3 as \left(-5x^{2}-x\right)+\left(-15x-3\right).

$$\left(-5x^{2}-x\right)+\left(-15x-3\right)$$

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

$$-x\left(5x+1\right)-3\left(5x+1\right)$$

Factor out common term 5x+1 by using distributive property.

$$\left(5x+1\right)\left(-x-3\right)$$

To find equation solutions, solve 5x+1=0 and -x-3=0.

$$x=-\frac{1}{5} x=-3$$

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}.

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

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

$$4x^{2}-4x+1=9x^{2}+12x+4$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+1-9x^{2}=12x+4$$

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

$$-5x^{2}-4x+1=12x+4$$

Subtract 12x from both sides.

$$-5x^{2}-4x+1-12x=4$$

Combine -4x and -12x to get -16x.

$$-5x^{2}-16x+1=4$$

Subtract 4 from both sides.

$$-5x^{2}-16x+1-4=0$$

Subtract 4 from 1 to get -3.

$$-5x^{2}-16x-3=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, -16 for b, and -3 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-16\right)±\sqrt{\left(-16\right)^{2}-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Square -16.

$$x=\frac{-\left(-16\right)±\sqrt{256-4\left(-5\right)\left(-3\right)}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-\left(-16\right)±\sqrt{256+20\left(-3\right)}}{2\left(-5\right)}$$

Multiply 20 times -3.

$$x=\frac{-\left(-16\right)±\sqrt{256-60}}{2\left(-5\right)}$$

Add 256 to -60.

$$x=\frac{-\left(-16\right)±\sqrt{196}}{2\left(-5\right)}$$

Take the square root of 196.

$$x=\frac{-\left(-16\right)±14}{2\left(-5\right)}$$

The opposite of -16 is 16.

$$x=\frac{16±14}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{16±14}{-10}$$

Now solve the equation x=\frac{16±14}{-10} when ± is plus. Add 16 to 14.

$$x=\frac{30}{-10}$$

Divide 30 by -10.

$$x=-3$$

Now solve the equation x=\frac{16±14}{-10} when ± is minus. Subtract 14 from 16.

$$x=\frac{2}{-10}$$

Reduce the fraction \frac{2}{-10} to lowest terms by extracting and canceling out 2.

$$x=-\frac{1}{5}$$

The equation is now solved.

$$x=-3 x=-\frac{1}{5}$$

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}.

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

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

$$4x^{2}-4x+1=9x^{2}+12x+4$$

Subtract 9x^{2} from both sides.

$$4x^{2}-4x+1-9x^{2}=12x+4$$

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

$$-5x^{2}-4x+1=12x+4$$

Subtract 12x from both sides.

$$-5x^{2}-4x+1-12x=4$$

Combine -4x and -12x to get -16x.

$$-5x^{2}-16x+1=4$$

Subtract 1 from both sides.

$$-5x^{2}-16x=4-1$$

Subtract 1 from 4 to get 3.

$$-5x^{2}-16x=3$$

Divide both sides by -5.

$$\frac{-5x^{2}-16x}{-5}=\frac{3}{-5}$$

Dividing by -5 undoes the multiplication by -5.

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

Divide -16 by -5.

$$x^{2}+\frac{16}{5}x=\frac{3}{-5}$$

Divide 3 by -5.

$$x^{2}+\frac{16}{5}x=-\frac{3}{5}$$

Divide \frac{16}{5}, the coefficient of the x term, by 2 to get \frac{8}{5}. Then add the square of \frac{8}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{16}{5}x+\left(\frac{8}{5}\right)^{2}=-\frac{3}{5}+\left(\frac{8}{5}\right)^{2}$$

Square \frac{8}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{16}{5}x+\frac{64}{25}=-\frac{3}{5}+\frac{64}{25}$$

Add -\frac{3}{5} to \frac{64}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{16}{5}x+\frac{64}{25}=\frac{49}{25}$$

Factor x^{2}+\frac{16}{5}x+\frac{64}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{8}{5}\right)^{2}=\frac{49}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{8}{5}\right)^{2}}=\sqrt{\frac{49}{25}}$$

Simplify.

$$x+\frac{8}{5}=\frac{7}{5} x+\frac{8}{5}=-\frac{7}{5}$$

Subtract \frac{8}{5} from both sides of the equation.

$$x=-\frac{1}{5} x=-3$$