Solve left(2x+3right)^2-left(3x-2right)^2=0

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

Solve for x

$$x=5$$

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

Steps Using Factoring By Grouping

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

$$4x^{2}+12x+9-\left(3x-2\right)^{2}=0$$

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

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

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

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

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

$$-5x^{2}+12x+9+12x-4=0$$

Combine 12x and 12x to get 24x.

$$-5x^{2}+24x+9-4=0$$

Subtract 4 from 9 to get 5.

$$-5x^{2}+24x+5=0$$

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

$$a+b=24 ab=-5\times 5=-25$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. List all such integer pairs that give product -25.

$$-1,25 -5,5$$

Calculate the sum for each pair.

$$-1+25=24 -5+5=0$$

The solution is the pair that gives sum 24.

$$a=25 b=-1$$

Rewrite -5x^{2}+24x+5 as \left(-5x^{2}+25x\right)+\left(-x+5\right).

$$\left(-5x^{2}+25x\right)+\left(-x+5\right)$$

Factor out 5x in -5x^{2}+25x.

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

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

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

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

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

Steps Using the Quadratic Formula

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

$$4x^{2}+12x+9-\left(3x-2\right)^{2}=0$$

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

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

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

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

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

$$-5x^{2}+12x+9+12x-4=0$$

Combine 12x and 12x to get 24x.

$$-5x^{2}+24x+9-4=0$$

Subtract 4 from 9 to get 5.

$$-5x^{2}+24x+5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -5 for a, 24 for b, and 5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-24±\sqrt{24^{2}-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Square 24.

$$x=\frac{-24±\sqrt{576-4\left(-5\right)\times 5}}{2\left(-5\right)}$$

Multiply -4 times -5.

$$x=\frac{-24±\sqrt{576+20\times 5}}{2\left(-5\right)}$$

Multiply 20 times 5.

$$x=\frac{-24±\sqrt{576+100}}{2\left(-5\right)}$$

Add 576 to 100.

$$x=\frac{-24±\sqrt{676}}{2\left(-5\right)}$$

Take the square root of 676.

$$x=\frac{-24±26}{2\left(-5\right)}$$

Multiply 2 times -5.

$$x=\frac{-24±26}{-10}$$

Now solve the equation x=\frac{-24±26}{-10} when ± is plus. Add -24 to 26.

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

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

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

Now solve the equation x=\frac{-24±26}{-10} when ± is minus. Subtract 26 from -24.

$$x=-\frac{50}{-10}$$

Divide -50 by -10.

$$x=5$$

The equation is now solved.

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

Steps for Completing the Square

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

$$4x^{2}+12x+9-\left(3x-2\right)^{2}=0$$

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

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

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

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

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

$$-5x^{2}+12x+9+12x-4=0$$

Combine 12x and 12x to get 24x.

$$-5x^{2}+24x+9-4=0$$

Subtract 4 from 9 to get 5.

$$-5x^{2}+24x+5=0$$

Subtract 5 from both sides. Anything subtracted from zero gives its negation.

$$-5x^{2}+24x=-5$$

Divide both sides by -5.

$$\frac{-5x^{2}+24x}{-5}=-\frac{5}{-5}$$

Dividing by -5 undoes the multiplication by -5.

$$x^{2}+\frac{24}{-5}x=-\frac{5}{-5}$$

Divide 24 by -5.

$$x^{2}-\frac{24}{5}x=-\frac{5}{-5}$$

Divide -5 by -5.

$$x^{2}-\frac{24}{5}x=1$$

Divide -\frac{24}{5}, the coefficient of the x term, by 2 to get -\frac{12}{5}. Then add the square of -\frac{12}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{24}{5}x+\left(-\frac{12}{5}\right)^{2}=1+\left(-\frac{12}{5}\right)^{2}$$

Square -\frac{12}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}=1+\frac{144}{25}$$

Add 1 to \frac{144}{25}.

$$x^{2}-\frac{24}{5}x+\frac{144}{25}=\frac{169}{25}$$

Factor x^{2}-\frac{24}{5}x+\frac{144}{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}.

$$\left(x-\frac{12}{5}\right)^{2}=\frac{169}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{12}{5}\right)^{2}}=\sqrt{\frac{169}{25}}$$

Simplify.

$$x-\frac{12}{5}=\frac{13}{5} x-\frac{12}{5}=-\frac{13}{5}$$

Add \frac{12}{5} to both sides of the equation.

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