Solve (6-2x)x=4

$$( 6 - 2 x ) x = 4$$

Solve for x

$$x=1$$

$$x=2$$

Steps Using the Quadratic Formula

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

$$6x-2x^{2}=4$$

Subtract 4 from both sides.

$$6x-2x^{2}-4=0$$

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.

$$-2x^{2}+6x-4=0$$

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

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

Square 6.

$$x=\frac{-6±\sqrt{36-4\left(-2\right)\left(-4\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-6±\sqrt{36+8\left(-4\right)}}{2\left(-2\right)}$$

Multiply 8 times -4.

$$x=\frac{-6±\sqrt{36-32}}{2\left(-2\right)}$$

Add 36 to -32.

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

Take the square root of 4.

$$x=\frac{-6±2}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-6±2}{-4}$$

Now solve the equation x=\frac{-6±2}{-4} when ± is plus. Add -6 to 2.

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

Divide -4 by -4.

$$x=1$$

Now solve the equation x=\frac{-6±2}{-4} when ± is minus. Subtract 2 from -6.

$$x=-\frac{8}{-4}$$

Divide -8 by -4.

$$x=2$$

The equation is now solved.

$$x=1 x=2$$

Steps for Completing the Square

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

$$6x-2x^{2}=4$$

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.

$$-2x^{2}+6x=4$$

Divide both sides by -2.

$$\frac{-2x^{2}+6x}{-2}=\frac{4}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{6}{-2}x=\frac{4}{-2}$$

Divide 6 by -2.

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

Divide 4 by -2.

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

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

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

Square -\frac{3}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-3x+\frac{9}{4}=-2+\frac{9}{4}$$

Add -2 to \frac{9}{4}.

$$x^{2}-3x+\frac{9}{4}=\frac{1}{4}$$

Factor x^{2}-3x+\frac{9}{4}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{3}{2} to both sides of the equation.

$$x=2 x=1$$