Solve (x-3)(3x-4)=12

$$(x-3)(3x-4)=12$$

Solve for x

$$x = \frac{13}{3} = 4\frac{1}{3} \approx 4.333333333$$

$$x=0$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x-3 by 3x-4 and combine like terms.

$$3x^{2}-13x+12=12$$

Subtract 12 from both sides.

$$3x^{2}-13x+12-12=0$$

Subtract 12 from 12 to get 0.

$$3x^{2}-13x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -13 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-13\right)±\sqrt{\left(-13\right)^{2}}}{2\times 3}$$

Take the square root of \left(-13\right)^{2}.

$$x=\frac{-\left(-13\right)±13}{2\times 3}$$

The opposite of -13 is 13.

$$x=\frac{13±13}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{13±13}{6}$$

Now solve the equation x=\frac{13±13}{6} when ± is plus. Add 13 to 13.

$$x=\frac{26}{6}$$

Reduce the fraction \frac{26}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{13}{3}$$

Now solve the equation x=\frac{13±13}{6} when ± is minus. Subtract 13 from 13.

$$x=\frac{0}{6}$$

Divide 0 by 6.

$$x=0$$

The equation is now solved.

$$x=\frac{13}{3} x=0$$

Steps for Completing the Square

Use the distributive property to multiply x-3 by 3x-4 and combine like terms.

$$3x^{2}-13x+12=12$$

Subtract 12 from both sides.

$$3x^{2}-13x=12-12$$

Subtract 12 from 12 to get 0.

$$3x^{2}-13x=0$$

Divide both sides by 3.

$$\frac{3x^{2}-13x}{3}=\frac{0}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{13}{3}x=\frac{0}{3}$$

Divide 0 by 3.

$$x^{2}-\frac{13}{3}x=0$$

Divide -\frac{13}{3}, the coefficient of the x term, by 2 to get -\frac{13}{6}. Then add the square of -\frac{13}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{13}{3}x+\left(-\frac{13}{6}\right)^{2}=\left(-\frac{13}{6}\right)^{2}$$

Square -\frac{13}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{13}{3}x+\frac{169}{36}=\frac{169}{36}$$

Factor x^{2}-\frac{13}{3}x+\frac{169}{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}.

$$\left(x-\frac{13}{6}\right)^{2}=\frac{169}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{13}{6}\right)^{2}}=\sqrt{\frac{169}{36}}$$

Simplify.

$$x-\frac{13}{6}=\frac{13}{6} x-\frac{13}{6}=-\frac{13}{6}$$

Add \frac{13}{6} to both sides of the equation.

$$x=\frac{13}{3} x=0$$