Solve (5x-2)^2=9

$$( 5 x - 2 ) ^ { 2 } = 9$$

Solve for x

$$x=1$$

$$x=-\frac{1}{5}=-0.2$$

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}.

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

Subtract 9 from both sides.

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

Subtract 9 from 4 to get -5.

$$25x^{2}-20x-5=0$$

Divide both sides by 5.

$$5x^{2}-4x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 5x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=-4 ab=5\left(-1\right)=-5$$

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. The only such pair is the system solution.

$$a=-5 b=1$$

Rewrite 5x^{2}-4x-1 as \left(5x^{2}-5x\right)+\left(x-1\right).

$$\left(5x^{2}-5x\right)+\left(x-1\right)$$

Factor out 5x in 5x^{2}-5x.

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

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

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

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

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

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}.

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

Subtract 9 from both sides.

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

Subtract 9 from 4 to get -5.

$$25x^{2}-20x-5=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 25 for a, -20 for b, and -5 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-20\right)±\sqrt{\left(-20\right)^{2}-4\times 25\left(-5\right)}}{2\times 25}$$

Square -20.

$$x=\frac{-\left(-20\right)±\sqrt{400-4\times 25\left(-5\right)}}{2\times 25}$$

Multiply -4 times 25.

$$x=\frac{-\left(-20\right)±\sqrt{400-100\left(-5\right)}}{2\times 25}$$

Multiply -100 times -5.

$$x=\frac{-\left(-20\right)±\sqrt{400+500}}{2\times 25}$$

Add 400 to 500.

$$x=\frac{-\left(-20\right)±\sqrt{900}}{2\times 25}$$

Take the square root of 900.

$$x=\frac{-\left(-20\right)±30}{2\times 25}$$

The opposite of -20 is 20.

$$x=\frac{20±30}{2\times 25}$$

Multiply 2 times 25.

$$x=\frac{20±30}{50}$$

Now solve the equation x=\frac{20±30}{50} when ± is plus. Add 20 to 30.

$$x=\frac{50}{50}$$

Divide 50 by 50.

$$x=1$$

Now solve the equation x=\frac{20±30}{50} when ± is minus. Subtract 30 from 20.

$$x=-\frac{10}{50}$$

Reduce the fraction \frac{-10}{50} to lowest terms by extracting and canceling out 10.

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

The equation is now solved.

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

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}.

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

Subtract 4 from both sides.

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

Subtract 4 from 9 to get 5.

$$25x^{2}-20x=5$$

Divide both sides by 25.

$$\frac{25x^{2}-20x}{25}=\frac{5}{25}$$

Dividing by 25 undoes the multiplication by 25.

$$x^{2}+\left(-\frac{20}{25}\right)x=\frac{5}{25}$$

Reduce the fraction \frac{-20}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{4}{5}x=\frac{5}{25}$$

Reduce the fraction \frac{5}{25} to lowest terms by extracting and canceling out 5.

$$x^{2}-\frac{4}{5}x=\frac{1}{5}$$

Divide -\frac{4}{5}, the coefficient of the x term, by 2 to get -\frac{2}{5}. Then add the square of -\frac{2}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{4}{5}x+\left(-\frac{2}{5}\right)^{2}=\frac{1}{5}+\left(-\frac{2}{5}\right)^{2}$$

Square -\frac{2}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{1}{5}+\frac{4}{25}$$

Add \frac{1}{5} to \frac{4}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{4}{5}x+\frac{4}{25}=\frac{9}{25}$$

Factor x^{2}-\frac{4}{5}x+\frac{4}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{2}{5}\right)^{2}=\frac{9}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{2}{5}\right)^{2}}=\sqrt{\frac{9}{25}}$$

Simplify.

$$x-\frac{2}{5}=\frac{3}{5} x-\frac{2}{5}=-\frac{3}{5}$$

Add \frac{2}{5} to both sides of the equation.

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