Solve 9(2x+3)^2-4(2x-5)^2=0

$$9 ( 2 x + 3 ) ^ { 2 } - 4 ( 2 x - 5 ) ^ { 2 } = 0$$

Solve for x

$$x = -\frac{19}{2} = -9\frac{1}{2} = -9.5$$

$$x=\frac{1}{10}=0.1$$

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}.

$$9\left(4x^{2}+12x+9\right)-4\left(2x-5\right)^{2}=0$$

Use the distributive property to multiply 9 by 4x^{2}+12x+9.

$$36x^{2}+108x+81-4\left(2x-5\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$36x^{2}+108x+81-4\left(4x^{2}-20x+25\right)=0$$

Use the distributive property to multiply -4 by 4x^{2}-20x+25.

$$36x^{2}+108x+81-16x^{2}+80x-100=0$$

Combine 36x^{2} and -16x^{2} to get 20x^{2}.

$$20x^{2}+108x+81+80x-100=0$$

Combine 108x and 80x to get 188x.

$$20x^{2}+188x+81-100=0$$

Subtract 100 from 81 to get -19.

$$20x^{2}+188x-19=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 20x^{2}+ax+bx-19. To find a and b, set up a system to be solved.

$$a+b=188 ab=20\left(-19\right)=-380$$

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 -380.

$$-1,380 -2,190 -4,95 -5,76 -10,38 -19,20$$

Calculate the sum for each pair.

$$-1+380=379 -2+190=188 -4+95=91 -5+76=71 -10+38=28 -19+20=1$$

The solution is the pair that gives sum 188.

$$a=-2 b=190$$

Rewrite 20x^{2}+188x-19 as \left(20x^{2}-2x\right)+\left(190x-19\right).

$$\left(20x^{2}-2x\right)+\left(190x-19\right)$$

Factor out 2x in the first and 19 in the second group.

$$2x\left(10x-1\right)+19\left(10x-1\right)$$

Factor out common term 10x-1 by using distributive property.

$$\left(10x-1\right)\left(2x+19\right)$$

To find equation solutions, solve 10x-1=0 and 2x+19=0.

$$x=\frac{1}{10} x=-\frac{19}{2}$$

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}.

$$9\left(4x^{2}+12x+9\right)-4\left(2x-5\right)^{2}=0$$

Use the distributive property to multiply 9 by 4x^{2}+12x+9.

$$36x^{2}+108x+81-4\left(2x-5\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$36x^{2}+108x+81-4\left(4x^{2}-20x+25\right)=0$$

Use the distributive property to multiply -4 by 4x^{2}-20x+25.

$$36x^{2}+108x+81-16x^{2}+80x-100=0$$

Combine 36x^{2} and -16x^{2} to get 20x^{2}.

$$20x^{2}+108x+81+80x-100=0$$

Combine 108x and 80x to get 188x.

$$20x^{2}+188x+81-100=0$$

Subtract 100 from 81 to get -19.

$$20x^{2}+188x-19=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 20 for a, 188 for b, and -19 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-188±\sqrt{188^{2}-4\times 20\left(-19\right)}}{2\times 20}$$

Square 188.

$$x=\frac{-188±\sqrt{35344-4\times 20\left(-19\right)}}{2\times 20}$$

Multiply -4 times 20.

$$x=\frac{-188±\sqrt{35344-80\left(-19\right)}}{2\times 20}$$

Multiply -80 times -19.

$$x=\frac{-188±\sqrt{35344+1520}}{2\times 20}$$

Add 35344 to 1520.

$$x=\frac{-188±\sqrt{36864}}{2\times 20}$$

Take the square root of 36864.

$$x=\frac{-188±192}{2\times 20}$$

Multiply 2 times 20.

$$x=\frac{-188±192}{40}$$

Now solve the equation x=\frac{-188±192}{40} when ± is plus. Add -188 to 192.

$$x=\frac{4}{40}$$

Reduce the fraction \frac{4}{40} to lowest terms by extracting and canceling out 4.

$$x=\frac{1}{10}$$

Now solve the equation x=\frac{-188±192}{40} when ± is minus. Subtract 192 from -188.

$$x=-\frac{380}{40}$$

Reduce the fraction \frac{-380}{40} to lowest terms by extracting and canceling out 20.

$$x=-\frac{19}{2}$$

The equation is now solved.

$$x=\frac{1}{10} x=-\frac{19}{2}$$

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}.

$$9\left(4x^{2}+12x+9\right)-4\left(2x-5\right)^{2}=0$$

Use the distributive property to multiply 9 by 4x^{2}+12x+9.

$$36x^{2}+108x+81-4\left(2x-5\right)^{2}=0$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2x-5\right)^{2}.

$$36x^{2}+108x+81-4\left(4x^{2}-20x+25\right)=0$$

Use the distributive property to multiply -4 by 4x^{2}-20x+25.

$$36x^{2}+108x+81-16x^{2}+80x-100=0$$

Combine 36x^{2} and -16x^{2} to get 20x^{2}.

$$20x^{2}+108x+81+80x-100=0$$

Combine 108x and 80x to get 188x.

$$20x^{2}+188x+81-100=0$$

Subtract 100 from 81 to get -19.

$$20x^{2}+188x-19=0$$

Add 19 to both sides. Anything plus zero gives itself.

$$20x^{2}+188x=19$$

Divide both sides by 20.

$$\frac{20x^{2}+188x}{20}=\frac{19}{20}$$

Dividing by 20 undoes the multiplication by 20.

$$x^{2}+\frac{188}{20}x=\frac{19}{20}$$

Reduce the fraction \frac{188}{20} to lowest terms by extracting and canceling out 4.

$$x^{2}+\frac{47}{5}x=\frac{19}{20}$$

Divide \frac{47}{5}, the coefficient of the x term, by 2 to get \frac{47}{10}. Then add the square of \frac{47}{10} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{47}{5}x+\left(\frac{47}{10}\right)^{2}=\frac{19}{20}+\left(\frac{47}{10}\right)^{2}$$

Square \frac{47}{10} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{47}{5}x+\frac{2209}{100}=\frac{19}{20}+\frac{2209}{100}$$

Add \frac{19}{20} to \frac{2209}{100} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{47}{5}x+\frac{2209}{100}=\frac{576}{25}$$

Factor x^{2}+\frac{47}{5}x+\frac{2209}{100}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{47}{10}\right)^{2}=\frac{576}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{47}{10}\right)^{2}}=\sqrt{\frac{576}{25}}$$

Simplify.

$$x+\frac{47}{10}=\frac{24}{5} x+\frac{47}{10}=-\frac{24}{5}$$

Subtract \frac{47}{10} from both sides of the equation.

$$x=\frac{1}{10} x=-\frac{19}{2}$$