Solve (6-x)2x=18

$$(6-x)2x=18$$

Solve for x

$$x=3$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 6-x by 2.

$$\left(12-2x\right)x=18$$

Use the distributive property to multiply 12-2x by x.

$$12x-2x^{2}=18$$

Subtract 18 from both sides.

$$12x-2x^{2}-18=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-2x^{2}+12x-18=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 12 for b, and -18 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-12±\sqrt{12^{2}-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Square 12.

$$x=\frac{-12±\sqrt{144-4\left(-2\right)\left(-18\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-12±\sqrt{144+8\left(-18\right)}}{2\left(-2\right)}$$

Multiply 8 times -18.

$$x=\frac{-12±\sqrt{144-144}}{2\left(-2\right)}$$

Add 144 to -144.

$$x=\frac{-12±\sqrt{0}}{2\left(-2\right)}$$

Take the square root of 0.

$$x=-\frac{12}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=-\frac{12}{-4}$$

Divide -12 by -4.

$$x=3$$

Steps for Completing the Square

Use the distributive property to multiply 6-x by 2.

$$\left(12-2x\right)x=18$$

Use the distributive property to multiply 12-2x by x.

$$12x-2x^{2}=18$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-2x^{2}+12x=18$$

Divide both sides by -2.

$$\frac{-2x^{2}+12x}{-2}=\frac{18}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{12}{-2}x=\frac{18}{-2}$$

Divide 12 by -2.

$$x^{2}-6x=\frac{18}{-2}$$

Divide 18 by -2.

$$x^{2}-6x=-9$$

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

$$x^{2}-6x+\left(-3\right)^{2}=-9+\left(-3\right)^{2}$$

Square -3.

$$x^{2}-6x+9=-9+9$$

Add -9 to 9.

$$x^{2}-6x+9=0$$

Factor x^{2}-6x+9. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-3\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-3\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-3=0 x-3=0$$

Add 3 to both sides of the equation.

$$x=3 x=3$$

The equation is now solved. Solutions are the same.

$$x=3$$