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

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

Solve for x

$$x=-\frac{1}{2}=-0.5$$

$$x=\frac{5}{8}=0.625$$

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}.

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

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

$$25x^{2}-20x+4-\left(9x^{2}-18x+9\right)=0$$

To find the opposite of 9x^{2}-18x+9, find the opposite of each term.

$$25x^{2}-20x+4-9x^{2}+18x-9=0$$

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

$$16x^{2}-20x+4+18x-9=0$$

Combine -20x and 18x to get -2x.

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

Subtract 9 from 4 to get -5.

$$16x^{2}-2x-5=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 16x^{2}+ax+bx-5. To find a and b, set up a system to be solved.

$$a+b=-2 ab=16\left(-5\right)=-80$$

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 -80.

$$1,-80 2,-40 4,-20 5,-16 8,-10$$

Calculate the sum for each pair.

$$1-80=-79 2-40=-38 4-20=-16 5-16=-11 8-10=-2$$

The solution is the pair that gives sum -2.

$$a=-10 b=8$$

Rewrite 16x^{2}-2x-5 as \left(16x^{2}-10x\right)+\left(8x-5\right).

$$\left(16x^{2}-10x\right)+\left(8x-5\right)$$

Factor out 2x in 16x^{2}-10x.

$$2x\left(8x-5\right)+8x-5$$

Factor out common term 8x-5 by using distributive property.

$$\left(8x-5\right)\left(2x+1\right)$$

To find equation solutions, solve 8x-5=0 and 2x+1=0.

$$x=\frac{5}{8} x=-\frac{1}{2}$$

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}.

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

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

$$25x^{2}-20x+4-\left(9x^{2}-18x+9\right)=0$$

To find the opposite of 9x^{2}-18x+9, find the opposite of each term.

$$25x^{2}-20x+4-9x^{2}+18x-9=0$$

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

$$16x^{2}-20x+4+18x-9=0$$

Combine -20x and 18x to get -2x.

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

Subtract 9 from 4 to get -5.

$$16x^{2}-2x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 16 for a, -2 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 16\left(-5\right)}}{2\times 16}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 16\left(-5\right)}}{2\times 16}$$

Multiply -4 times 16.

$$x=\frac{-\left(-2\right)±\sqrt{4-64\left(-5\right)}}{2\times 16}$$

Multiply -64 times -5.

$$x=\frac{-\left(-2\right)±\sqrt{4+320}}{2\times 16}$$

Add 4 to 320.

$$x=\frac{-\left(-2\right)±\sqrt{324}}{2\times 16}$$

Take the square root of 324.

$$x=\frac{-\left(-2\right)±18}{2\times 16}$$

The opposite of -2 is 2.

$$x=\frac{2±18}{2\times 16}$$

Multiply 2 times 16.

$$x=\frac{2±18}{32}$$

Now solve the equation x=\frac{2±18}{32} when ± is plus. Add 2 to 18.

$$x=\frac{20}{32}$$

Reduce the fraction \frac{20}{32} to lowest terms by extracting and canceling out 4.

$$x=\frac{5}{8}$$

Now solve the equation x=\frac{2±18}{32} when ± is minus. Subtract 18 from 2.

$$x=-\frac{16}{32}$$

Reduce the fraction \frac{-16}{32} to lowest terms by extracting and canceling out 16.

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

The equation is now solved.

$$x=\frac{5}{8} x=-\frac{1}{2}$$

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}.

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

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

$$25x^{2}-20x+4-\left(9x^{2}-18x+9\right)=0$$

To find the opposite of 9x^{2}-18x+9, find the opposite of each term.

$$25x^{2}-20x+4-9x^{2}+18x-9=0$$

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

$$16x^{2}-20x+4+18x-9=0$$

Combine -20x and 18x to get -2x.

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

Subtract 9 from 4 to get -5.

$$16x^{2}-2x-5=0$$

Add 5 to both sides. Anything plus zero gives itself.

$$16x^{2}-2x=5$$

Divide both sides by 16.

$$\frac{16x^{2}-2x}{16}=\frac{5}{16}$$

Dividing by 16 undoes the multiplication by 16.

$$x^{2}+\left(-\frac{2}{16}\right)x=\frac{5}{16}$$

Reduce the fraction \frac{-2}{16} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{1}{8}x=\frac{5}{16}$$

Divide -\frac{1}{8}, the coefficient of the x term, by 2 to get -\frac{1}{16}. Then add the square of -\frac{1}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{8}x+\left(-\frac{1}{16}\right)^{2}=\frac{5}{16}+\left(-\frac{1}{16}\right)^{2}$$

Square -\frac{1}{16} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{5}{16}+\frac{1}{256}$$

Add \frac{5}{16} to \frac{1}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{8}x+\frac{1}{256}=\frac{81}{256}$$

Factor x^{2}-\frac{1}{8}x+\frac{1}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{16}\right)^{2}=\frac{81}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{16}\right)^{2}}=\sqrt{\frac{81}{256}}$$

Simplify.

$$x-\frac{1}{16}=\frac{9}{16} x-\frac{1}{16}=-\frac{9}{16}$$

Add \frac{1}{16} to both sides of the equation.

$$x=\frac{5}{8} x=-\frac{1}{2}$$