Solve (x+1)^2-(2x-3)^2=0

$$( x + 1 ) ^ { 2 } - ( 2 x - 3 ) ^ { 2 } = 0$$

Solve for x

$$x=4$$

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

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}.

$$x^{2}+2x+1-\left(2x-3\right)^{2}=0$$

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

$$x^{2}+2x+1-\left(4x^{2}-12x+9\right)=0$$

To find the opposite of 4x^{2}-12x+9, find the opposite of each term.

$$x^{2}+2x+1-4x^{2}+12x-9=0$$

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

$$-3x^{2}+2x+1+12x-9=0$$

Combine 2x and 12x to get 14x.

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

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 positive, a and b are both positive. List all such integer pairs that give product 24.

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

Calculate the sum for each pair.

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

The solution is the pair that gives sum 14.

$$a=12 b=2$$

Rewrite -3x^{2}+14x-8 as \left(-3x^{2}+12x\right)+\left(2x-8\right).

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

Factor out 3x in the first and -2 in the second group.

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

Factor out common term -x+4 by using distributive property.

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

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

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

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}.

$$x^{2}+2x+1-\left(2x-3\right)^{2}=0$$

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

$$x^{2}+2x+1-\left(4x^{2}-12x+9\right)=0$$

To find the opposite of 4x^{2}-12x+9, find the opposite of each term.

$$x^{2}+2x+1-4x^{2}+12x-9=0$$

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

$$-3x^{2}+2x+1+12x-9=0$$

Combine 2x and 12x to get 14x.

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

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}.

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

Square 14.

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

Multiply -4 times -3.

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

Multiply 12 times -8.

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

Add 196 to -96.

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

Take the square root of 100.

$$x=\frac{-14±10}{2\left(-3\right)}$$

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.

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

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

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

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

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

Divide -24 by -6.

$$x=4$$

The equation is now solved.

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

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}.

$$x^{2}+2x+1-\left(2x-3\right)^{2}=0$$

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

$$x^{2}+2x+1-\left(4x^{2}-12x+9\right)=0$$

To find the opposite of 4x^{2}-12x+9, find the opposite of each term.

$$x^{2}+2x+1-4x^{2}+12x-9=0$$

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

$$-3x^{2}+2x+1+12x-9=0$$

Combine 2x and 12x to get 14x.

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

Subtract 9 from 1 to get -8.

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

Add 8 to both sides. Anything plus zero gives itself.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

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.

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

Add \frac{7}{3} to both sides of the equation.

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