Solve (9-x)(x)=20

$$( 9 - x ) ( x ) = 20$$

Solve for x

$$x=4$$

$$x=5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 9-x by x.

$$9x-x^{2}=20$$

Subtract 20 from both sides.

$$9x-x^{2}-20=0$$

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.

$$-x^{2}+9x-20=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -1 for a, 9 for b, and -20 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-9±\sqrt{9^{2}-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Square 9.

$$x=\frac{-9±\sqrt{81-4\left(-1\right)\left(-20\right)}}{2\left(-1\right)}$$

Multiply -4 times -1.

$$x=\frac{-9±\sqrt{81+4\left(-20\right)}}{2\left(-1\right)}$$

Multiply 4 times -20.

$$x=\frac{-9±\sqrt{81-80}}{2\left(-1\right)}$$

Add 81 to -80.

$$x=\frac{-9±\sqrt{1}}{2\left(-1\right)}$$

Take the square root of 1.

$$x=\frac{-9±1}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-9±1}{-2}$$

Now solve the equation x=\frac{-9±1}{-2} when ± is plus. Add -9 to 1.

$$x=-\frac{8}{-2}$$

Divide -8 by -2.

$$x=4$$

Now solve the equation x=\frac{-9±1}{-2} when ± is minus. Subtract 1 from -9.

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

Divide -10 by -2.

$$x=5$$

The equation is now solved.

$$x=4 x=5$$

Steps for Completing the Square

Use the distributive property to multiply 9-x by x.

$$9x-x^{2}=20$$

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.

$$-x^{2}+9x=20$$

Divide both sides by -1.

$$\frac{-x^{2}+9x}{-1}=\frac{20}{-1}$$

Dividing by -1 undoes the multiplication by -1.

$$x^{2}+\frac{9}{-1}x=\frac{20}{-1}$$

Divide 9 by -1.

$$x^{2}-9x=\frac{20}{-1}$$

Divide 20 by -1.

$$x^{2}-9x=-20$$

Divide -9, the coefficient of the x term, by 2 to get -\frac{9}{2}. Then add the square of -\frac{9}{2} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-9x+\left(-\frac{9}{2}\right)^{2}=-20+\left(-\frac{9}{2}\right)^{2}$$

Square -\frac{9}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-9x+\frac{81}{4}=-20+\frac{81}{4}$$

Add -20 to \frac{81}{4}.

$$x^{2}-9x+\frac{81}{4}=\frac{1}{4}$$

Factor x^{2}-9x+\frac{81}{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}.

$$\left(x-\frac{9}{2}\right)^{2}=\frac{1}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{9}{2}\right)^{2}}=\sqrt{\frac{1}{4}}$$

Simplify.

$$x-\frac{9}{2}=\frac{1}{2} x-\frac{9}{2}=-\frac{1}{2}$$

Add \frac{9}{2} to both sides of the equation.

$$x=5 x=4$$