Solve 9x^2+36x+36=22-9x

$$9 { x }^{ 2 } +36x+36=22-9x$$

Solve for x

$$x = -\frac{14}{3} = -4\frac{2}{3} \approx -4.666666667$$

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

Steps Using Factoring By Grouping

Subtract 22 from both sides.

$$9x^{2}+36x+36-22=-9x$$

Subtract 22 from 36 to get 14.

$$9x^{2}+36x+14=-9x$$

Add 9x to both sides.

$$9x^{2}+36x+14+9x=0$$

Combine 36x and 9x to get 45x.

$$9x^{2}+45x+14=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as 9x^{2}+ax+bx+14. To find a and b, set up a system to be solved.

$$a+b=45 ab=9\times 14=126$$

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 126.

$$1,126 2,63 3,42 6,21 7,18 9,14$$

Calculate the sum for each pair.

$$1+126=127 2+63=65 3+42=45 6+21=27 7+18=25 9+14=23$$

The solution is the pair that gives sum 45.

$$a=3 b=42$$

Rewrite 9x^{2}+45x+14 as \left(9x^{2}+3x\right)+\left(42x+14\right).

$$\left(9x^{2}+3x\right)+\left(42x+14\right)$$

Factor out 3x in the first and 14 in the second group.

$$3x\left(3x+1\right)+14\left(3x+1\right)$$

Factor out common term 3x+1 by using distributive property.

$$\left(3x+1\right)\left(3x+14\right)$$

To find equation solutions, solve 3x+1=0 and 3x+14=0.

$$x=-\frac{1}{3} x=-\frac{14}{3}$$

Steps Using the Quadratic Formula

Subtract 22 from both sides.

$$9x^{2}+36x+36-22=-9x$$

Subtract 22 from 36 to get 14.

$$9x^{2}+36x+14=-9x$$

Add 9x to both sides.

$$9x^{2}+36x+14+9x=0$$

Combine 36x and 9x to get 45x.

$$9x^{2}+45x+14=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 45 for b, and 14 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-45±\sqrt{45^{2}-4\times 9\times 14}}{2\times 9}$$

Square 45.

$$x=\frac{-45±\sqrt{2025-4\times 9\times 14}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-45±\sqrt{2025-36\times 14}}{2\times 9}$$

Multiply -36 times 14.

$$x=\frac{-45±\sqrt{2025-504}}{2\times 9}$$

Add 2025 to -504.

$$x=\frac{-45±\sqrt{1521}}{2\times 9}$$

Take the square root of 1521.

$$x=\frac{-45±39}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-45±39}{18}$$

Now solve the equation x=\frac{-45±39}{18} when ± is plus. Add -45 to 39.

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

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

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

Now solve the equation x=\frac{-45±39}{18} when ± is minus. Subtract 39 from -45.

$$x=-\frac{84}{18}$$

Reduce the fraction \frac{-84}{18} to lowest terms by extracting and canceling out 6.

$$x=-\frac{14}{3}$$

The equation is now solved.

$$x=-\frac{1}{3} x=-\frac{14}{3}$$

Steps for Completing the Square

Add 9x to both sides.

$$9x^{2}+36x+36+9x=22$$

Combine 36x and 9x to get 45x.

$$9x^{2}+45x+36=22$$

Subtract 36 from both sides.

$$9x^{2}+45x=22-36$$

Subtract 36 from 22 to get -14.

$$9x^{2}+45x=-14$$

Divide both sides by 9.

$$\frac{9x^{2}+45x}{9}=-\frac{14}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{45}{9}x=-\frac{14}{9}$$

Divide 45 by 9.

$$x^{2}+5x=-\frac{14}{9}$$

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.

$$x^{2}+5x+\left(\frac{5}{2}\right)^{2}=-\frac{14}{9}+\left(\frac{5}{2}\right)^{2}$$

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

$$x^{2}+5x+\frac{25}{4}=-\frac{14}{9}+\frac{25}{4}$$

Add -\frac{14}{9} to \frac{25}{4} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+5x+\frac{25}{4}=\frac{169}{36}$$

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}.

$$\left(x+\frac{5}{2}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{5}{2}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x+\frac{5}{2}=\frac{13}{6} x+\frac{5}{2}=-\frac{13}{6}$$

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

$$x=-\frac{1}{3} x=-\frac{14}{3}$$