Solve (3x+1)(x+2)=1

$$( 3 x + 1 ) ( x + 2 ) = 1$$

Solve for x

$$x=\frac{\sqrt{37}-7}{6}\approx -0.152872912$$

$$x=\frac{-\sqrt{37}-7}{6}\approx -2.180460422$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x+1 by x+2 and combine like terms.

$$3x^{2}+7x+2=1$$

Subtract 1 from both sides.

$$3x^{2}+7x+2-1=0$$

Subtract 1 from 2 to get 1.

$$3x^{2}+7x+1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 7 for b, and 1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-7±\sqrt{7^{2}-4\times 3}}{2\times 3}$$

Square 7.

$$x=\frac{-7±\sqrt{49-4\times 3}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-7±\sqrt{49-12}}{2\times 3}$$

Add 49 to -12.

$$x=\frac{-7±\sqrt{37}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-7±\sqrt{37}}{6}$$

Now solve the equation x=\frac{-7±\sqrt{37}}{6} when ± is plus. Add -7 to \sqrt{37}.

$$x=\frac{\sqrt{37}-7}{6}$$

Now solve the equation x=\frac{-7±\sqrt{37}}{6} when ± is minus. Subtract \sqrt{37} from -7.

$$x=\frac{-\sqrt{37}-7}{6}$$

The equation is now solved.

$$x=\frac{\sqrt{37}-7}{6} x=\frac{-\sqrt{37}-7}{6}$$

Steps for Completing the Square

Use the distributive property to multiply 3x+1 by x+2 and combine like terms.

$$3x^{2}+7x+2=1$$

Subtract 2 from both sides.

$$3x^{2}+7x=1-2$$

Subtract 2 from 1 to get -1.

$$3x^{2}+7x=-1$$

Divide both sides by 3.

$$\frac{3x^{2}+7x}{3}=-\frac{1}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{7}{3}x=-\frac{1}{3}$$

Divide \frac{7}{3}, the coefficient of the x term, by 2 to get \frac{7}{6}. Then add the square of \frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{7}{3}x+\left(\frac{7}{6}\right)^{2}=-\frac{1}{3}+\left(\frac{7}{6}\right)^{2}$$

Square \frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=-\frac{1}{3}+\frac{49}{36}$$

Add -\frac{1}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{7}{3}x+\frac{49}{36}=\frac{37}{36}$$

Factor x^{2}+\frac{7}{3}x+\frac{49}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{7}{6}\right)^{2}=\frac{37}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{7}{6}\right)^{2}}=\sqrt{\frac{37}{36}}$$

Simplify.

$$x+\frac{7}{6}=\frac{\sqrt{37}}{6} x+\frac{7}{6}=-\frac{\sqrt{37}}{6}$$

Subtract \frac{7}{6} from both sides of the equation.

$$x=\frac{\sqrt{37}-7}{6} x=\frac{-\sqrt{37}-7}{6}$$