Solve x(3x-1)-3=1

$$x(3x-1)-3=1$$

Solve for x

$$x = \frac{4}{3} = 1\frac{1}{3} \approx 1.333333333$$

$$x=-1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 3x-1.

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

Subtract 1 from both sides.

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

Subtract 1 from -3 to get -4.

$$3x^{2}-x-4=0$$

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

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 3\left(-4\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-1\right)±\sqrt{1-12\left(-4\right)}}{2\times 3}$$

Multiply -12 times -4.

$$x=\frac{-\left(-1\right)±\sqrt{1+48}}{2\times 3}$$

Add 1 to 48.

$$x=\frac{-\left(-1\right)±\sqrt{49}}{2\times 3}$$

Take the square root of 49.

$$x=\frac{-\left(-1\right)±7}{2\times 3}$$

The opposite of -1 is 1.

$$x=\frac{1±7}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{1±7}{6}$$

Now solve the equation x=\frac{1±7}{6} when ± is plus. Add 1 to 7.

$$x=\frac{8}{6}$$

Reduce the fraction \frac{8}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{4}{3}$$

Now solve the equation x=\frac{1±7}{6} when ± is minus. Subtract 7 from 1.

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

Divide -6 by 6.

$$x=-1$$

The equation is now solved.

$$x=\frac{4}{3} x=-1$$

Steps for Completing the Square

Use the distributive property to multiply x by 3x-1.

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

Add 3 to both sides.

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

Add 1 and 3 to get 4.

$$3x^{2}-x=4$$

Divide both sides by 3.

$$\frac{3x^{2}-x}{3}=\frac{4}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{1}{3}x=\frac{4}{3}$$

Divide -\frac{1}{3}, the coefficient of the x term, by 2 to get -\frac{1}{6}. Then add the square of -\frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{3}x+\left(-\frac{1}{6}\right)^{2}=\frac{4}{3}+\left(-\frac{1}{6}\right)^{2}$$

Square -\frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{3}x+\frac{1}{36}=\frac{4}{3}+\frac{1}{36}$$

Add \frac{4}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

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

Factor x^{2}-\frac{1}{3}x+\frac{1}{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}.

$$\left(x-\frac{1}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{1}{6}=\frac{7}{6} x-\frac{1}{6}=-\frac{7}{6}$$

Add \frac{1}{6} to both sides of the equation.

$$x=\frac{4}{3} x=-1$$