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

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

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-1\right)^{2}.

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

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

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

Subtract 4x^{2} from both sides.

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

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

$$-3x^{2}-2x+1=12x+9$$

Subtract 12x from both sides.

$$-3x^{2}-2x+1-12x=9$$

Combine -2x and -12x to get -14x.

$$-3x^{2}-14x+1=9$$

Subtract 9 from both sides.

$$-3x^{2}-14x+1-9=0$$

Subtract 9 from 1 to get -8.

$$-3x^{2}-14x-8=0$$

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.

$$a+b=-14 ab=-3\left(-8\right)=24$$

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 24.

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

Calculate the sum for each pair.

$$-1-24=-25 -2-12=-14 -3-8=-11 -4-6=-10$$

The solution is the pair that gives sum -14.

$$a=-2 b=-12$$

Rewrite -3x^{2}-14x-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-1\right)^{2}.

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

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

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

Subtract 4x^{2} from both sides.

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

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

$$-3x^{2}-2x+1=12x+9$$

Subtract 12x from both sides.

$$-3x^{2}-2x+1-12x=9$$

Combine -2x and -12x to get -14x.

$$-3x^{2}-14x+1=9$$

Subtract 9 from both sides.

$$-3x^{2}-14x+1-9=0$$

Subtract 9 from 1 to get -8.

$$-3x^{2}-14x-8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, -14 for b, and -8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-14\right)±\sqrt{\left(-14\right)^{2}-4\left(-3\right)\left(-8\right)}}{2\left(-3\right)}$$

Square -14.

$$x=\frac{-\left(-14\right)±\sqrt{196-4\left(-3\right)\left(-8\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-\left(-14\right)±\sqrt{196+12\left(-8\right)}}{2\left(-3\right)}$$

Multiply 12 times -8.

$$x=\frac{-\left(-14\right)±\sqrt{196-96}}{2\left(-3\right)}$$

Add 196 to -96.

$$x=\frac{-\left(-14\right)±\sqrt{100}}{2\left(-3\right)}$$

Take the square root of 100.

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

The opposite of -14 is 14.

$$x=\frac{14±10}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{14±10}{-6}$$

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

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

Divide 24 by -6.

$$x=-4$$

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

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

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-1\right)^{2}.

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

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

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

Subtract 4x^{2} from both sides.

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

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

$$-3x^{2}-2x+1=12x+9$$

Subtract 12x from both sides.

$$-3x^{2}-2x+1-12x=9$$

Combine -2x and -12x to get -14x.

$$-3x^{2}-14x+1=9$$

Subtract 1 from both sides.

$$-3x^{2}-14x=9-1$$

Subtract 1 from 9 to get 8.

$$-3x^{2}-14x=8$$

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide -14 by -3.

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

Divide 8 by -3.

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

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.

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

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

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

Add -\frac{8}{3} to \frac{49}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{14}{3}x+\frac{49}{9}=\frac{25}{9}$$

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}.

$$\left(x+\frac{7}{3}\right)^{2}=\frac{25}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{3}\right)^{2}}=\sqrt{\frac{25}{9}}$$

Simplify.

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

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

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