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

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

Solve for x

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

$$x = \frac{7}{2} = 3\frac{1}{2} = 3.5$$

Steps Using Factoring By Grouping

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

$$9x^{2}-24x+16-\left(x+3\right)^{2}=0$$

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

$$9x^{2}-24x+16-\left(x^{2}+6x+9\right)=0$$

To find the opposite of x^{2}+6x+9, find the opposite of each term.

$$9x^{2}-24x+16-x^{2}-6x-9=0$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-24x+16-6x-9=0$$

Combine -24x and -6x to get -30x.

$$8x^{2}-30x+16-9=0$$

Subtract 9 from 16 to get 7.

$$8x^{2}-30x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 8x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=-30 ab=8\times 7=56$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 56.

$$-1,-56 -2,-28 -4,-14 -7,-8$$

Calculate the sum for each pair.

$$-1-56=-57 -2-28=-30 -4-14=-18 -7-8=-15$$

The solution is the pair that gives sum -30.

$$a=-28 b=-2$$

Rewrite 8x^{2}-30x+7 as \left(8x^{2}-28x\right)+\left(-2x+7\right).

$$\left(8x^{2}-28x\right)+\left(-2x+7\right)$$

Factor out 4x in the first and -1 in the second group.

$$4x\left(2x-7\right)-\left(2x-7\right)$$

Factor out common term 2x-7 by using distributive property.

$$\left(2x-7\right)\left(4x-1\right)$$

To find equation solutions, solve 2x-7=0 and 4x-1=0.

$$x=\frac{7}{2} x=\frac{1}{4}$$

Steps Using the Quadratic Formula

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

$$9x^{2}-24x+16-\left(x+3\right)^{2}=0$$

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

$$9x^{2}-24x+16-\left(x^{2}+6x+9\right)=0$$

To find the opposite of x^{2}+6x+9, find the opposite of each term.

$$9x^{2}-24x+16-x^{2}-6x-9=0$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-24x+16-6x-9=0$$

Combine -24x and -6x to get -30x.

$$8x^{2}-30x+16-9=0$$

Subtract 9 from 16 to get 7.

$$8x^{2}-30x+7=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 8 for a, -30 for b, and 7 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 8\times 7}}{2\times 8}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 8\times 7}}{2\times 8}$$

Multiply -4 times 8.

$$x=\frac{-\left(-30\right)±\sqrt{900-32\times 7}}{2\times 8}$$

Multiply -32 times 7.

$$x=\frac{-\left(-30\right)±\sqrt{900-224}}{2\times 8}$$

Add 900 to -224.

$$x=\frac{-\left(-30\right)±\sqrt{676}}{2\times 8}$$

Take the square root of 676.

$$x=\frac{-\left(-30\right)±26}{2\times 8}$$

The opposite of -30 is 30.

$$x=\frac{30±26}{2\times 8}$$

Multiply 2 times 8.

$$x=\frac{30±26}{16}$$

Now solve the equation x=\frac{30±26}{16} when ± is plus. Add 30 to 26.

$$x=\frac{56}{16}$$

Reduce the fraction \frac{56}{16} to lowest terms by extracting and canceling out 8.

$$x=\frac{7}{2}$$

Now solve the equation x=\frac{30±26}{16} when ± is minus. Subtract 26 from 30.

$$x=\frac{4}{16}$$

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

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

The equation is now solved.

$$x=\frac{7}{2} x=\frac{1}{4}$$

Steps for Completing the Square

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

$$9x^{2}-24x+16-\left(x+3\right)^{2}=0$$

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

$$9x^{2}-24x+16-\left(x^{2}+6x+9\right)=0$$

To find the opposite of x^{2}+6x+9, find the opposite of each term.

$$9x^{2}-24x+16-x^{2}-6x-9=0$$

Combine 9x^{2} and -x^{2} to get 8x^{2}.

$$8x^{2}-24x+16-6x-9=0$$

Combine -24x and -6x to get -30x.

$$8x^{2}-30x+16-9=0$$

Subtract 9 from 16 to get 7.

$$8x^{2}-30x+7=0$$

Subtract 7 from both sides. Anything subtracted from zero gives its negation.

$$8x^{2}-30x=-7$$

Divide both sides by 8.

$$\frac{8x^{2}-30x}{8}=-\frac{7}{8}$$

Dividing by 8 undoes the multiplication by 8.

$$x^{2}+\left(-\frac{30}{8}\right)x=-\frac{7}{8}$$

Reduce the fraction \frac{-30}{8} to lowest terms by extracting and canceling out 2.

$$x^{2}-\frac{15}{4}x=-\frac{7}{8}$$

Divide -\frac{15}{4}, the coefficient of the x term, by 2 to get -\frac{15}{8}. Then add the square of -\frac{15}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{15}{4}x+\left(-\frac{15}{8}\right)^{2}=-\frac{7}{8}+\left(-\frac{15}{8}\right)^{2}$$

Square -\frac{15}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{15}{4}x+\frac{225}{64}=-\frac{7}{8}+\frac{225}{64}$$

Add -\frac{7}{8} to \frac{225}{64} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{15}{4}x+\frac{225}{64}=\frac{169}{64}$$

Factor x^{2}-\frac{15}{4}x+\frac{225}{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}.

$$\left(x-\frac{15}{8}\right)^{2}=\frac{169}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{15}{8}\right)^{2}}=\sqrt{\frac{169}{64}}$$

Simplify.

$$x-\frac{15}{8}=\frac{13}{8} x-\frac{15}{8}=-\frac{13}{8}$$

Add \frac{15}{8} to both sides of the equation.

$$x=\frac{7}{2} x=\frac{1}{4}$$