Solve (3x+5)^2=4

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

Solve for x

$$x=-1$$

$$x = -\frac{7}{3} = -2\frac{1}{3} \approx -2.333333333$$

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}.

$$9x^{2}+30x+25=4$$

Subtract 4 from both sides.

$$9x^{2}+30x+25-4=0$$

Subtract 4 from 25 to get 21.

$$9x^{2}+30x+21=0$$

Divide both sides by 3.

$$3x^{2}+10x+7=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 3x^{2}+ax+bx+7. To find a and b, set up a system to be solved.

$$a+b=10 ab=3\times 7=21$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 21.

$$1,21 3,7$$

Calculate the sum for each pair.

$$1+21=22 3+7=10$$

The solution is the pair that gives sum 10.

$$a=3 b=7$$

Rewrite 3x^{2}+10x+7 as \left(3x^{2}+3x\right)+\left(7x+7\right).

$$\left(3x^{2}+3x\right)+\left(7x+7\right)$$

Factor out 3x in the first and 7 in the second group.

$$3x\left(x+1\right)+7\left(x+1\right)$$

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

$$\left(x+1\right)\left(3x+7\right)$$

To find equation solutions, solve x+1=0 and 3x+7=0.

$$x=-1 x=-\frac{7}{3}$$

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}.

$$9x^{2}+30x+25=4$$

Subtract 4 from both sides.

$$9x^{2}+30x+25-4=0$$

Subtract 4 from 25 to get 21.

$$9x^{2}+30x+21=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 30 for b, and 21 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-30±\sqrt{30^{2}-4\times 9\times 21}}{2\times 9}$$

Square 30.

$$x=\frac{-30±\sqrt{900-4\times 9\times 21}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-30±\sqrt{900-36\times 21}}{2\times 9}$$

Multiply -36 times 21.

$$x=\frac{-30±\sqrt{900-756}}{2\times 9}$$

Add 900 to -756.

$$x=\frac{-30±\sqrt{144}}{2\times 9}$$

Take the square root of 144.

$$x=\frac{-30±12}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-30±12}{18}$$

Now solve the equation x=\frac{-30±12}{18} when ± is plus. Add -30 to 12.

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

Divide -18 by 18.

$$x=-1$$

Now solve the equation x=\frac{-30±12}{18} when ± is minus. Subtract 12 from -30.

$$x=-\frac{42}{18}$$

Reduce the fraction \frac{-42}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{7}{3}$$

The equation is now solved.

$$x=-1 x=-\frac{7}{3}$$

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}.

$$9x^{2}+30x+25=4$$

Subtract 25 from both sides.

$$9x^{2}+30x=4-25$$

Subtract 25 from 4 to get -21.

$$9x^{2}+30x=-21$$

Divide both sides by 9.

$$\frac{9x^{2}+30x}{9}=-\frac{21}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{30}{9}x=-\frac{21}{9}$$

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

$$x^{2}+\frac{10}{3}x=-\frac{21}{9}$$

Reduce the fraction \frac{-21}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}+\frac{10}{3}x=-\frac{7}{3}$$

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.

$$x^{2}+\frac{10}{3}x+\left(\frac{5}{3}\right)^{2}=-\frac{7}{3}+\left(\frac{5}{3}\right)^{2}$$

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

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=-\frac{7}{3}+\frac{25}{9}$$

Add -\frac{7}{3} to \frac{25}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{10}{3}x+\frac{25}{9}=\frac{4}{9}$$

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}.

$$\left(x+\frac{5}{3}\right)^{2}=\frac{4}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{3}\right)^{2}}=\sqrt{\frac{4}{9}}$$

Simplify.

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

Subtract \frac{5}{3} from both sides of the equation.

$$x=-1 x=-\frac{7}{3}$$