Solve (2x+4)3x-12=x

$$(2x+4)3x-12=x$$

Solve for x

$$x=\frac{\sqrt{409}-11}{12}\approx 0.768645701$$

$$x=\frac{-\sqrt{409}-11}{12}\approx -2.601979035$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2x+4 by 3.

$$\left(6x+12\right)x-12=x$$

Use the distributive property to multiply 6x+12 by x.

$$6x^{2}+12x-12=x$$

Subtract x from both sides.

$$6x^{2}+12x-12-x=0$$

Combine 12x and -x to get 11x.

$$6x^{2}+11x-12=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, 11 for b, and -12 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\times 6\left(-12\right)}}{2\times 6}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\times 6\left(-12\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-11±\sqrt{121-24\left(-12\right)}}{2\times 6}$$

Multiply -24 times -12.

$$x=\frac{-11±\sqrt{121+288}}{2\times 6}$$

Add 121 to 288.

$$x=\frac{-11±\sqrt{409}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{-11±\sqrt{409}}{12}$$

Now solve the equation x=\frac{-11±\sqrt{409}}{12} when ± is plus. Add -11 to \sqrt{409}.

$$x=\frac{\sqrt{409}-11}{12}$$

Now solve the equation x=\frac{-11±\sqrt{409}}{12} when ± is minus. Subtract \sqrt{409} from -11.

$$x=\frac{-\sqrt{409}-11}{12}$$

The equation is now solved.

$$x=\frac{\sqrt{409}-11}{12} x=\frac{-\sqrt{409}-11}{12}$$

Steps for Completing the Square

Use the distributive property to multiply 2x+4 by 3.

$$\left(6x+12\right)x-12=x$$

Use the distributive property to multiply 6x+12 by x.

$$6x^{2}+12x-12=x$$

Subtract x from both sides.

$$6x^{2}+12x-12-x=0$$

Combine 12x and -x to get 11x.

$$6x^{2}+11x-12=0$$

Add 12 to both sides. Anything plus zero gives itself.

$$6x^{2}+11x=12$$

Divide both sides by 6.

$$\frac{6x^{2}+11x}{6}=\frac{12}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}+\frac{11}{6}x=\frac{12}{6}$$

Divide 12 by 6.

$$x^{2}+\frac{11}{6}x=2$$

Divide \frac{11}{6}, the coefficient of the x term, by 2 to get \frac{11}{12}. Then add the square of \frac{11}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{11}{6}x+\left(\frac{11}{12}\right)^{2}=2+\left(\frac{11}{12}\right)^{2}$$

Square \frac{11}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{11}{6}x+\frac{121}{144}=2+\frac{121}{144}$$

Add 2 to \frac{121}{144}.

$$x^{2}+\frac{11}{6}x+\frac{121}{144}=\frac{409}{144}$$

Factor x^{2}+\frac{11}{6}x+\frac{121}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{11}{12}\right)^{2}=\frac{409}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{11}{12}\right)^{2}}=\sqrt{\frac{409}{144}}$$

Simplify.

$$x+\frac{11}{12}=\frac{\sqrt{409}}{12} x+\frac{11}{12}=-\frac{\sqrt{409}}{12}$$

Subtract \frac{11}{12} from both sides of the equation.

$$x=\frac{\sqrt{409}-11}{12} x=\frac{-\sqrt{409}-11}{12}$$