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

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

Solve for x

$$x=\frac{2}{3}\approx 0.666666667$$

$$x=-4$$

Steps Using Factoring By Grouping

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

$$x^{2}-6x+9=\left(2x+1\right)^{2}$$

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

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

Subtract 4x^{2} from both sides.

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

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

$$-3x^{2}-6x+9=4x+1$$

Subtract 4x from both sides.

$$-3x^{2}-6x+9-4x=1$$

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

$$-3x^{2}-10x+9=1$$

Subtract 1 from both sides.

$$-3x^{2}-10x+9-1=0$$

Subtract 1 from 9 to get 8.

$$-3x^{2}-10x+8=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx+8. To find a and b, set up a system to be solved.

$$a+b=-10 ab=-3\times 8=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$a=2 b=-12$$

Rewrite -3x^{2}-10x+8 as \left(-3x^{2}+2x\right)+\left(-12x+8\right).

$$\left(-3x^{2}+2x\right)+\left(-12x+8\right)$$

Factor out -x in the first and -4 in the second group.

$$-x\left(3x-2\right)-4\left(3x-2\right)$$

Factor out common term 3x-2 by using distributive property.

$$\left(3x-2\right)\left(-x-4\right)$$

To find equation solutions, solve 3x-2=0 and -x-4=0.

$$x=\frac{2}{3} x=-4$$

Steps Using the Quadratic Formula

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

$$x^{2}-6x+9=\left(2x+1\right)^{2}$$

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

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

Subtract 4x^{2} from both sides.

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

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

$$-3x^{2}-6x+9=4x+1$$

Subtract 4x from both sides.

$$-3x^{2}-6x+9-4x=1$$

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

$$-3x^{2}-10x+9=1$$

Subtract 1 from both sides.

$$-3x^{2}-10x+9-1=0$$

Subtract 1 from 9 to get 8.

$$-3x^{2}-10x+8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -10 for b, and 8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-3\right)\times 8}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-10\right)±\sqrt{100+12\times 8}}{2\left(-3\right)}$$

Multiply 12 times 8.

$$x=\frac{-\left(-10\right)±\sqrt{100+96}}{2\left(-3\right)}$$

Add 100 to 96.

$$x=\frac{-\left(-10\right)±\sqrt{196}}{2\left(-3\right)}$$

Take the square root of 196.

$$x=\frac{-\left(-10\right)±14}{2\left(-3\right)}$$

The opposite of -10 is 10.

$$x=\frac{10±14}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{10±14}{-6}$$

Now solve the equation x=\frac{10±14}{-6} when ± is plus. Add 10 to 14.

$$x=\frac{24}{-6}$$

Divide 24 by -6.

$$x=-4$$

Now solve the equation x=\frac{10±14}{-6} when ± is minus. Subtract 14 from 10.

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

Reduce the fraction \frac{-4}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{3}$$

The equation is now solved.

$$x=-4 x=\frac{2}{3}$$

Steps for Completing the Square

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

$$x^{2}-6x+9=\left(2x+1\right)^{2}$$

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

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

Subtract 4x^{2} from both sides.

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

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

$$-3x^{2}-6x+9=4x+1$$

Subtract 4x from both sides.

$$-3x^{2}-6x+9-4x=1$$

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

$$-3x^{2}-10x+9=1$$

Subtract 9 from both sides.

$$-3x^{2}-10x=1-9$$

Subtract 9 from 1 to get -8.

$$-3x^{2}-10x=-8$$

Divide both sides by -3.

$$\frac{-3x^{2}-10x}{-3}=-\frac{8}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\left(-\frac{10}{-3}\right)x=-\frac{8}{-3}$$

Divide -10 by -3.

$$x^{2}+\frac{10}{3}x=-\frac{8}{-3}$$

Divide -8 by -3.

$$x^{2}+\frac{10}{3}x=\frac{8}{3}$$

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.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=\frac{8}{3}+\left(\frac{5}{3}\right)^{2}$$

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

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{8}{3}+\frac{25}{9}$$

Add \frac{8}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{49}{9}$$

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}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x+\frac{5}{3}=\frac{7}{3} x+\frac{5}{3}=-\frac{7}{3}$$

Subtract \frac{5}{3} from both sides of the equation.

$$x=\frac{2}{3} x=-4$$