Solve left(3x+1right)^2=-2x

$${ \left(3x+1 \right) }^{ 2 } =-2x$$

Solve for x

$$x=\frac{\sqrt{7}-4}{9}\approx -0.150472077$$

$$x=\frac{-\sqrt{7}-4}{9}\approx -0.738416812$$

Steps Using the Quadratic Formula

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$9x^{2}+6x+1=-2x$$

Add 2x to both sides.

$$9x^{2}+6x+1+2x=0$$

Combine 6x and 2x to get 8x.

$$9x^{2}+8x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, 8 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 9}}{2\times 9}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 9}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-8±\sqrt{64-36}}{2\times 9}$$

Add 64 to -36.

$$x=\frac{-8±\sqrt{28}}{2\times 9}$$

Take the square root of 28.

$$x=\frac{-8±2\sqrt{7}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{-8±2\sqrt{7}}{18}$$

Now solve the equation x=\frac{-8±2\sqrt{7}}{18} when ± is plus. Add -8 to 2\sqrt{7}.

$$x=\frac{2\sqrt{7}-8}{18}$$

Divide -8+2\sqrt{7} by 18.

$$x=\frac{\sqrt{7}-4}{9}$$

Now solve the equation x=\frac{-8±2\sqrt{7}}{18} when ± is minus. Subtract 2\sqrt{7} from -8.

$$x=\frac{-2\sqrt{7}-8}{18}$$

Divide -8-2\sqrt{7} by 18.

$$x=\frac{-\sqrt{7}-4}{9}$$

The equation is now solved.

$$x=\frac{\sqrt{7}-4}{9} x=\frac{-\sqrt{7}-4}{9}$$

Steps for Completing the Square

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(3x+1\right)^{2}.

$$9x^{2}+6x+1=-2x$$

Add 2x to both sides.

$$9x^{2}+6x+1+2x=0$$

Combine 6x and 2x to get 8x.

$$9x^{2}+8x+1=0$$

Subtract 1 from both sides. Anything subtracted from zero gives its negation.

$$9x^{2}+8x=-1$$

Divide both sides by 9.

$$\frac{9x^{2}+8x}{9}=-\frac{1}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\frac{8}{9}x=-\frac{1}{9}$$

Divide \frac{8}{9}, the coefficient of the x term, by 2 to get \frac{4}{9}. Then add the square of \frac{4}{9} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{8}{9}x+\left(\frac{4}{9}\right)^{2}=-\frac{1}{9}+\left(\frac{4}{9}\right)^{2}$$

Square \frac{4}{9} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{8}{9}x+\frac{16}{81}=-\frac{1}{9}+\frac{16}{81}$$

Add -\frac{1}{9} to \frac{16}{81} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{8}{9}x+\frac{16}{81}=\frac{7}{81}$$

Factor x^{2}+\frac{8}{9}x+\frac{16}{81}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{4}{9}\right)^{2}=\frac{7}{81}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{4}{9}\right)^{2}}=\sqrt{\frac{7}{81}}$$

Simplify.

$$x+\frac{4}{9}=\frac{\sqrt{7}}{9} x+\frac{4}{9}=-\frac{\sqrt{7}}{9}$$

Subtract \frac{4}{9} from both sides of the equation.

$$x=\frac{\sqrt{7}-4}{9} x=\frac{-\sqrt{7}-4}{9}$$