Solve (3a+5)^2=0

$$( 3 a + 5 ) ^ { 2 } = 0$$

Solve for a

$$a = -\frac{5}{3} = -1\frac{2}{3} \approx -1.666666667$$

Steps Using Factoring By Grouping

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

$$9a^{2}+30a+25=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9a^{2}+aa+ba+25. To find a and b, set up a system to be solved.

$$a+b=30 ab=9\times 25=225$$

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 225.

$$1,225 3,75 5,45 9,25 15,15$$

Calculate the sum for each pair.

$$1+225=226 3+75=78 5+45=50 9+25=34 15+15=30$$

The solution is the pair that gives sum 30.

$$a=15 b=15$$

Rewrite 9a^{2}+30a+25 as \left(9a^{2}+15a\right)+\left(15a+25\right).

$$\left(9a^{2}+15a\right)+\left(15a+25\right)$$

Factor out 3a in the first and 5 in the second group.

$$3a\left(3a+5\right)+5\left(3a+5\right)$$

Factor out common term 3a+5 by using distributive property.

$$\left(3a+5\right)\left(3a+5\right)$$

Rewrite as a binomial square.

$$\left(3a+5\right)^{2}$$

To find equation solution, solve 3a+5=0.

$$a=-\frac{5}{3}$$

Steps Using the Quadratic Formula

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

$$9a^{2}+30a+25=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 30 for b, and 25 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$a=\frac{-30±\sqrt{30^{2}-4\times 9\times 25}}{2\times 9}$$

Square 30.

$$a=\frac{-30±\sqrt{900-4\times 9\times 25}}{2\times 9}$$

Multiply -4 times 9.

$$a=\frac{-30±\sqrt{900-36\times 25}}{2\times 9}$$

Multiply -36 times 25.

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

Add 900 to -900.

$$a=\frac{-30±\sqrt{0}}{2\times 9}$$

Take the square root of 0.

$$a=-\frac{30}{2\times 9}$$

Multiply 2 times 9.

$$a=-\frac{30}{18}$$

Reduce the fraction \frac{-30}{18} to lowest terms by extracting and canceling out 6.

$$a=-\frac{5}{3}$$

Steps for Completing the Square

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

$$9a^{2}+30a+25=0$$

Subtract 25 from both sides. Anything subtracted from zero gives its negation.

$$9a^{2}+30a=-25$$

Divide both sides by 9.

$$\frac{9a^{2}+30a}{9}=-\frac{25}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$a^{2}+\frac{30}{9}a=-\frac{25}{9}$$

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

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

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.

$$a^{2}+\frac{10}{3}a+\left(\frac{5}{3}\right)^{2}=-\frac{25}{9}+\left(\frac{5}{3}\right)^{2}$$

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

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

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

$$a^{2}+\frac{10}{3}a+\frac{25}{9}=0$$

Factor a^{2}+\frac{10}{3}a+\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}.

$$\left(a+\frac{5}{3}\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(a+\frac{5}{3}\right)^{2}}=\sqrt{0}$$

Simplify.

$$a+\frac{5}{3}=0 a+\frac{5}{3}=0$$

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

$$a=-\frac{5}{3} a=-\frac{5}{3}$$

The equation is now solved. Solutions are the same.

$$a=-\frac{5}{3}$$