Solve x*(33-3x)=72

$$x \times (33-3x)=72$$

Solve for x

$$x=3$$

$$x=8$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by 33-3x.

$$33x-3x^{2}=72$$

Subtract 72 from both sides.

$$33x-3x^{2}-72=0$$

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.

$$-3x^{2}+33x-72=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 33 for b, and -72 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-33±\sqrt{33^{2}-4\left(-3\right)\left(-72\right)}}{2\left(-3\right)}$$

Square 33.

$$x=\frac{-33±\sqrt{1089-4\left(-3\right)\left(-72\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-33±\sqrt{1089+12\left(-72\right)}}{2\left(-3\right)}$$

Multiply 12 times -72.

$$x=\frac{-33±\sqrt{1089-864}}{2\left(-3\right)}$$

Add 1089 to -864.

$$x=\frac{-33±\sqrt{225}}{2\left(-3\right)}$$

Take the square root of 225.

$$x=\frac{-33±15}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-33±15}{-6}$$

Now solve the equation x=\frac{-33±15}{-6} when ± is plus. Add -33 to 15.

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

Divide -18 by -6.

$$x=3$$

Now solve the equation x=\frac{-33±15}{-6} when ± is minus. Subtract 15 from -33.

$$x=-\frac{48}{-6}$$

Divide -48 by -6.

$$x=8$$

The equation is now solved.

$$x=3 x=8$$

Steps for Completing the Square

Use the distributive property to multiply x by 33-3x.

$$33x-3x^{2}=72$$

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.

$$-3x^{2}+33x=72$$

Divide both sides by -3.

$$\frac{-3x^{2}+33x}{-3}=\frac{72}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{33}{-3}x=\frac{72}{-3}$$

Divide 33 by -3.

$$x^{2}-11x=\frac{72}{-3}$$

Divide 72 by -3.

$$x^{2}-11x=-24$$

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

$$x^{2}-11x+\left(-\frac{11}{2}\right)^{2}=-24+\left(-\frac{11}{2}\right)^{2}$$

Square -\frac{11}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-11x+\frac{121}{4}=-24+\frac{121}{4}$$

Add -24 to \frac{121}{4}.

$$x^{2}-11x+\frac{121}{4}=\frac{25}{4}$$

Factor x^{2}-11x+\frac{121}{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}.

$$\left(x-\frac{11}{2}\right)^{2}=\frac{25}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{2}\right)^{2}}=\sqrt{\frac{25}{4}}$$

Simplify.

$$x-\frac{11}{2}=\frac{5}{2} x-\frac{11}{2}=-\frac{5}{2}$$

Add \frac{11}{2} to both sides of the equation.

$$x=8 x=3$$