Solve (20-2x)x=42

$$(20-2x)x=42$$

Solve for x

$$x=3$$

$$x=7$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 20-2x by x.

$$20x-2x^{2}=42$$

Subtract 42 from both sides.

$$20x-2x^{2}-42=0$$

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.

$$-2x^{2}+20x-42=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -2 for a, 20 for b, and -42 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-20±\sqrt{20^{2}-4\left(-2\right)\left(-42\right)}}{2\left(-2\right)}$$

Square 20.

$$x=\frac{-20±\sqrt{400-4\left(-2\right)\left(-42\right)}}{2\left(-2\right)}$$

Multiply -4 times -2.

$$x=\frac{-20±\sqrt{400+8\left(-42\right)}}{2\left(-2\right)}$$

Multiply 8 times -42.

$$x=\frac{-20±\sqrt{400-336}}{2\left(-2\right)}$$

Add 400 to -336.

$$x=\frac{-20±\sqrt{64}}{2\left(-2\right)}$$

Take the square root of 64.

$$x=\frac{-20±8}{2\left(-2\right)}$$

Multiply 2 times -2.

$$x=\frac{-20±8}{-4}$$

Now solve the equation x=\frac{-20±8}{-4} when ± is plus. Add -20 to 8.

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

Divide -12 by -4.

$$x=3$$

Now solve the equation x=\frac{-20±8}{-4} when ± is minus. Subtract 8 from -20.

$$x=-\frac{28}{-4}$$

Divide -28 by -4.

$$x=7$$

The equation is now solved.

$$x=3 x=7$$

Steps for Completing the Square

Use the distributive property to multiply 20-2x by x.

$$20x-2x^{2}=42$$

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.

$$-2x^{2}+20x=42$$

Divide both sides by -2.

$$\frac{-2x^{2}+20x}{-2}=\frac{42}{-2}$$

Dividing by -2 undoes the multiplication by -2.

$$x^{2}+\frac{20}{-2}x=\frac{42}{-2}$$

Divide 20 by -2.

$$x^{2}-10x=\frac{42}{-2}$$

Divide 42 by -2.

$$x^{2}-10x=-21$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=-21+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=-21+25$$

Add -21 to 25.

$$x^{2}-10x+25=4$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=4$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{4}$$

Simplify.

$$x-5=2 x-5=-2$$

Add 5 to both sides of the equation.

$$x=7 x=3$$