Solve (3x-5)^2-16=0

$$( 3 x - 5 ) ^ { 2 } - 16 = 0$$

Solve for x

$$x=3$$

$$x=\frac{1}{3}\approx 0.333333333$$

Steps Using Factoring By Grouping

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-16=0$$

Subtract 16 from 25 to get 9.

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

Divide both sides by 3.

$$3x^{2}-10x+3=0$$

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

$$a+b=-10 ab=3\times 3=9$$

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 9.

$$-1,-9 -3,-3$$

Calculate the sum for each pair.

$$-1-9=-10 -3-3=-6$$

The solution is the pair that gives sum -10.

$$a=-9 b=-1$$

Rewrite 3x^{2}-10x+3 as \left(3x^{2}-9x\right)+\left(-x+3\right).

$$\left(3x^{2}-9x\right)+\left(-x+3\right)$$

Factor out 3x in the first and -1 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-16=0$$

Subtract 16 from 25 to get 9.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -30 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-30\right)±\sqrt{\left(-30\right)^{2}-4\times 9\times 9}}{2\times 9}$$

Square -30.

$$x=\frac{-\left(-30\right)±\sqrt{900-4\times 9\times 9}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-30\right)±\sqrt{900-36\times 9}}{2\times 9}$$

Multiply -36 times 9.

$$x=\frac{-\left(-30\right)±\sqrt{900-324}}{2\times 9}$$

Add 900 to -324.

$$x=\frac{-\left(-30\right)±\sqrt{576}}{2\times 9}$$

Take the square root of 576.

$$x=\frac{-\left(-30\right)±24}{2\times 9}$$

The opposite of -30 is 30.

$$x=\frac{30±24}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{30±24}{18}$$

Now solve the equation x=\frac{30±24}{18} when ± is plus. Add 30 to 24.

$$x=\frac{54}{18}$$

Divide 54 by 18.

$$x=3$$

Now solve the equation x=\frac{30±24}{18} when ± is minus. Subtract 24 from 30.

$$x=\frac{6}{18}$$

Reduce the fraction \frac{6}{18} to lowest terms by extracting and canceling out 6.

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

The equation is now solved.

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

Steps for Completing the Square

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3x-5\right)^{2}.

$$9x^{2}-30x+25-16=0$$

Subtract 16 from 25 to get 9.

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

Subtract 9 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}-30x=-9$$

Divide both sides by 9.

$$\frac{9x^{2}-30x}{9}=-\frac{9}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{30}{9}\right)x=-\frac{9}{9}$$

Reduce the fraction \frac{-30}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{10}{3}x=-\frac{9}{9}$$

Divide -9 by 9.

$$x^{2}-\frac{10}{3}x=-1$$

Divide -\frac{10}{3}, the coefficient of the x term, by 2 to get -\frac{5}{3}. Then add the square of -\frac{5}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{10}{3}x+\left(-\frac{5}{3}\right)^{2}=-1+\left(-\frac{5}{3}\right)^{2}$$

Square -\frac{5}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=-1+\frac{25}{9}$$

Add -1 to \frac{25}{9}.

$$x^{2}-\frac{10}{3}x+\frac{25}{9}=\frac{16}{9}$$

Factor x^{2}-\frac{10}{3}x+\frac{25}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{3}\right)^{2}=\frac{16}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{3}\right)^{2}}=\sqrt{\frac{16}{9}}$$

Simplify.

$$x-\frac{5}{3}=\frac{4}{3} x-\frac{5}{3}=-\frac{4}{3}$$

Add \frac{5}{3} to both sides of the equation.

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