Solve left(3x+5right)^2+5left(3x+5right)+6=0

$${ \left(3x+5 \right) }^{ 2 } +5 \left( 3x+5 \right) +6=0$$

Solve for x

$$x = -\frac{8}{3} = -2\frac{2}{3} \approx -2.666666667$$

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

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}.

$$9x^{2}+30x+25+5\left(3x+5\right)+6=0$$

Use the distributive property to multiply 5 by 3x+5.

$$9x^{2}+30x+25+15x+25+6=0$$

Combine 30x and 15x to get 45x.

$$9x^{2}+45x+25+25+6=0$$

Add 25 and 25 to get 50.

$$9x^{2}+45x+50+6=0$$

Add 50 and 6 to get 56.

$$9x^{2}+45x+56=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+56. To find a and b, set up a system to be solved.

$$a+b=45 ab=9\times 56=504$$

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 504.

$$1,504 2,252 3,168 4,126 6,84 7,72 8,63 9,56 12,42 14,36 18,28 21,24$$

Calculate the sum for each pair.

$$1+504=505 2+252=254 3+168=171 4+126=130 6+84=90 7+72=79 8+63=71 9+56=65 12+42=54 14+36=50 18+28=46 21+24=45$$

The solution is the pair that gives sum 45.

$$a=21 b=24$$

Rewrite 9x^{2}+45x+56 as \left(9x^{2}+21x\right)+\left(24x+56\right).

$$\left(9x^{2}+21x\right)+\left(24x+56\right)$$

Factor out 3x in the first and 8 in the second group.

$$3x\left(3x+7\right)+8\left(3x+7\right)$$

Factor out common term 3x+7 by using distributive property.

$$\left(3x+7\right)\left(3x+8\right)$$

To find equation solutions, solve 3x+7=0 and 3x+8=0.

$$x=-\frac{7}{3} x=-\frac{8}{3}$$

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}.

$$9x^{2}+30x+25+5\left(3x+5\right)+6=0$$

Use the distributive property to multiply 5 by 3x+5.

$$9x^{2}+30x+25+15x+25+6=0$$

Combine 30x and 15x to get 45x.

$$9x^{2}+45x+25+25+6=0$$

Add 25 and 25 to get 50.

$$9x^{2}+45x+50+6=0$$

Add 50 and 6 to get 56.

$$9x^{2}+45x+56=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 45 for b, and 56 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-45±\sqrt{45^{2}-4\times 9\times 56}}{2\times 9}$$

Square 45.

$$x=\frac{-45±\sqrt{2025-4\times 9\times 56}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-45±\sqrt{2025-36\times 56}}{2\times 9}$$

Multiply -36 times 56.

$$x=\frac{-45±\sqrt{2025-2016}}{2\times 9}$$

Add 2025 to -2016.

$$x=\frac{-45±\sqrt{9}}{2\times 9}$$

Take the square root of 9.

$$x=\frac{-45±3}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-45±3}{18}$$

Now solve the equation x=\frac{-45±3}{18} when ± is plus. Add -45 to 3.

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

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

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

Now solve the equation x=\frac{-45±3}{18} when ± is minus. Subtract 3 from -45.

$$x=-\frac{48}{18}$$

Reduce the fraction \frac{-48}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{8}{3}$$

The equation is now solved.

$$x=-\frac{7}{3} x=-\frac{8}{3}$$

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}.

$$9x^{2}+30x+25+5\left(3x+5\right)+6=0$$

Use the distributive property to multiply 5 by 3x+5.

$$9x^{2}+30x+25+15x+25+6=0$$

Combine 30x and 15x to get 45x.

$$9x^{2}+45x+25+25+6=0$$

Add 25 and 25 to get 50.

$$9x^{2}+45x+50+6=0$$

Add 50 and 6 to get 56.

$$9x^{2}+45x+56=0$$

Subtract 56 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}+45x=-56$$

Divide both sides by 9.

$$\frac{9x^{2}+45x}{9}=-\frac{56}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{45}{9}x=-\frac{56}{9}$$

Divide 45 by 9.

$$x^{2}+5x=-\frac{56}{9}$$

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

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{56}{9}+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=-\frac{56}{9}+\frac{25}{4}$$

Add -\frac{56}{9} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+5x+\frac{25}{4}=\frac{1}{36}$$

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

$$\left(x+\frac{5}{2}\right)^{2}=\frac{1}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{1}{36}}$$

Simplify.

$$x+\frac{5}{2}=\frac{1}{6} x+\frac{5}{2}=-\frac{1}{6}$$

Subtract \frac{5}{2} from both sides of the equation.

$$x=-\frac{7}{3} x=-\frac{8}{3}$$