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

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

Solve for x

$$x=-\frac{1}{4}=-0.25$$

$$x=3$$

Steps Using Factoring By Grouping

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

$$25x^{2}-20x+4-\left(3x+4\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+4\right)^{2}.

$$25x^{2}-20x+4-\left(9x^{2}+24x+16\right)=0$$

To find the opposite of 9x^{2}+24x+16, find the opposite of each term.

$$25x^{2}-20x+4-9x^{2}-24x-16=0$$

Combine 25x^{2} and -9x^{2} to get 16x^{2}.

$$16x^{2}-20x+4-24x-16=0$$

Combine -20x and -24x to get -44x.

$$16x^{2}-44x+4-16=0$$

Subtract 16 from 4 to get -12.

$$16x^{2}-44x-12=0$$

Divide both sides by 4.

$$4x^{2}-11x-3=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 4x^{2}+ax+bx-3. To find a and b, set up a system to be solved.

$$a+b=-11 ab=4\left(-3\right)=-12$$

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 -12.

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

Calculate the sum for each pair.

$$1-12=-11 2-6=-4 3-4=-1$$

The solution is the pair that gives sum -11.

$$a=-12 b=1$$

Rewrite 4x^{2}-11x-3 as \left(4x^{2}-12x\right)+\left(x-3\right).

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

Factor out 4x in 4x^{2}-12x.

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

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

$$\left(x-3\right)\left(4x+1\right)$$

To find equation solutions, solve x-3=0 and 4x+1=0.

$$x=3 x=-\frac{1}{4}$$

Steps Using the Quadratic Formula

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

$$25x^{2}-20x+4-\left(3x+4\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+4\right)^{2}.

$$25x^{2}-20x+4-\left(9x^{2}+24x+16\right)=0$$

To find the opposite of 9x^{2}+24x+16, find the opposite of each term.

$$25x^{2}-20x+4-9x^{2}-24x-16=0$$

Combine 25x^{2} and -9x^{2} to get 16x^{2}.

$$16x^{2}-20x+4-24x-16=0$$

Combine -20x and -24x to get -44x.

$$16x^{2}-44x+4-16=0$$

Subtract 16 from 4 to get -12.

$$16x^{2}-44x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -44 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-44\right)±\sqrt{\left(-44\right)^{2}-4\times 16\left(-12\right)}}{2\times 16}$$

Square -44.

$$x=\frac{-\left(-44\right)±\sqrt{1936-4\times 16\left(-12\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-44\right)±\sqrt{1936-64\left(-12\right)}}{2\times 16}$$

Multiply -64 times -12.

$$x=\frac{-\left(-44\right)±\sqrt{1936+768}}{2\times 16}$$

Add 1936 to 768.

$$x=\frac{-\left(-44\right)±\sqrt{2704}}{2\times 16}$$

Take the square root of 2704.

$$x=\frac{-\left(-44\right)±52}{2\times 16}$$

The opposite of -44 is 44.

$$x=\frac{44±52}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{44±52}{32}$$

Now solve the equation x=\frac{44±52}{32} when ± is plus. Add 44 to 52.

$$x=\frac{96}{32}$$

Divide 96 by 32.

$$x=3$$

Now solve the equation x=\frac{44±52}{32} when ± is minus. Subtract 52 from 44.

$$x=-\frac{8}{32}$$

Reduce the fraction \frac{-8}{32} to lowest terms by extracting and canceling out 8.

$$x=-\frac{1}{4}$$

The equation is now solved.

$$x=3 x=-\frac{1}{4}$$

Steps for Completing the Square

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

$$25x^{2}-20x+4-\left(3x+4\right)^{2}=0$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+4\right)^{2}.

$$25x^{2}-20x+4-\left(9x^{2}+24x+16\right)=0$$

To find the opposite of 9x^{2}+24x+16, find the opposite of each term.

$$25x^{2}-20x+4-9x^{2}-24x-16=0$$

Combine 25x^{2} and -9x^{2} to get 16x^{2}.

$$16x^{2}-20x+4-24x-16=0$$

Combine -20x and -24x to get -44x.

$$16x^{2}-44x+4-16=0$$

Subtract 16 from 4 to get -12.

$$16x^{2}-44x-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$16x^{2}-44x=12$$

Divide both sides by 16.

$$\frac{16x^{2}-44x}{16}=\frac{12}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{44}{16}\right)x=\frac{12}{16}$$

Reduce the fraction \frac{-44}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{11}{4}x=\frac{12}{16}$$

Reduce the fraction \frac{12}{16} to lowest terms by extracting and canceling out 4.

$$x^{2}-\frac{11}{4}x=\frac{3}{4}$$

Divide -\frac{11}{4}, the coefficient of the x term, by 2 to get -\frac{11}{8}. Then add the square of -\frac{11}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{4}x+\left(-\frac{11}{8}\right)^{2}=\frac{3}{4}+\left(-\frac{11}{8}\right)^{2}$$

Square -\frac{11}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=\frac{3}{4}+\frac{121}{64}$$

Add \frac{3}{4} to \frac{121}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{4}x+\frac{121}{64}=\frac{169}{64}$$

Factor x^{2}-\frac{11}{4}x+\frac{121}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{8}\right)^{2}=\frac{169}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{8}\right)^{2}}=\sqrt{\frac{169}{64}}$$

Simplify.

$$x-\frac{11}{8}=\frac{13}{8} x-\frac{11}{8}=-\frac{13}{8}$$

Add \frac{11}{8} to both sides of the equation.

$$x=3 x=-\frac{1}{4}$$