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

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

Solve for x

$$x = \frac{19}{4} = 4\frac{3}{4} = 4.75$$

$$x=0$$

Steps Using the Quadratic Formula

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

$$4x^{2}-19x+12=12$$

Subtract 12 from both sides.

$$4x^{2}-19x+12-12=0$$

Subtract 12 from 12 to get 0.

$$4x^{2}-19x=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 4 for a, -19 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-19\right)±\sqrt{\left(-19\right)^{2}}}{2\times 4}$$

Take the square root of \left(-19\right)^{2}.

$$x=\frac{-\left(-19\right)±19}{2\times 4}$$

The opposite of -19 is 19.

$$x=\frac{19±19}{2\times 4}$$

Multiply 2 times 4.

$$x=\frac{19±19}{8}$$

Now solve the equation x=\frac{19±19}{8} when ± is plus. Add 19 to 19.

$$x=\frac{38}{8}$$

Reduce the fraction \frac{38}{8} to lowest terms by extracting and canceling out 2.

$$x=\frac{19}{4}$$

Now solve the equation x=\frac{19±19}{8} when ± is minus. Subtract 19 from 19.

$$x=\frac{0}{8}$$

Divide 0 by 8.

$$x=0$$

The equation is now solved.

$$x=\frac{19}{4} x=0$$

Steps for Completing the Square

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

$$4x^{2}-19x+12=12$$

Subtract 12 from both sides.

$$4x^{2}-19x=12-12$$

Subtract 12 from 12 to get 0.

$$4x^{2}-19x=0$$

Divide both sides by 4.

$$\frac{4x^{2}-19x}{4}=\frac{0}{4}$$

Dividing by 4 undoes the multiplication by 4.

$$x^{2}-\frac{19}{4}x=\frac{0}{4}$$

Divide 0 by 4.

$$x^{2}-\frac{19}{4}x=0$$

Divide -\frac{19}{4}, the coefficient of the x term, by 2 to get -\frac{19}{8}. Then add the square of -\frac{19}{8} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{19}{4}x+\left(-\frac{19}{8}\right)^{2}=\left(-\frac{19}{8}\right)^{2}$$

Square -\frac{19}{8} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{19}{4}x+\frac{361}{64}=\frac{361}{64}$$

Factor x^{2}-\frac{19}{4}x+\frac{361}{64}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{19}{8}\right)^{2}=\frac{361}{64}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{19}{8}\right)^{2}}=\sqrt{\frac{361}{64}}$$

Simplify.

$$x-\frac{19}{8}=\frac{19}{8} x-\frac{19}{8}=-\frac{19}{8}$$

Add \frac{19}{8} to both sides of the equation.

$$x=\frac{19}{4} x=0$$