Solve 3x(11-2x)=15

$$3x(11-2x)=15$$

Solve for x

$$x=5$$

$$x=\frac{1}{2}=0.5$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 3x by 11-2x.

$$33x-6x^{2}=15$$

Subtract 15 from both sides.

$$33x-6x^{2}-15=0$$

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.

$$-6x^{2}+33x-15=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 33 for b, and -15 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-33±\sqrt{33^{2}-4\left(-6\right)\left(-15\right)}}{2\left(-6\right)}$$

Square 33.

$$x=\frac{-33±\sqrt{1089-4\left(-6\right)\left(-15\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-33±\sqrt{1089+24\left(-15\right)}}{2\left(-6\right)}$$

Multiply 24 times -15.

$$x=\frac{-33±\sqrt{1089-360}}{2\left(-6\right)}$$

Add 1089 to -360.

$$x=\frac{-33±\sqrt{729}}{2\left(-6\right)}$$

Take the square root of 729.

$$x=\frac{-33±27}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-33±27}{-12}$$

Now solve the equation x=\frac{-33±27}{-12} when ± is plus. Add -33 to 27.

$$x=-\frac{6}{-12}$$

Reduce the fraction \frac{-6}{-12} to lowest terms by extracting and canceling out 6.

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

Now solve the equation x=\frac{-33±27}{-12} when ± is minus. Subtract 27 from -33.

$$x=-\frac{60}{-12}$$

Divide -60 by -12.

$$x=5$$

The equation is now solved.

$$x=\frac{1}{2} x=5$$

Steps for Completing the Square

Use the distributive property to multiply 3x by 11-2x.

$$33x-6x^{2}=15$$

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.

$$-6x^{2}+33x=15$$

Divide both sides by -6.

$$\frac{-6x^{2}+33x}{-6}=\frac{15}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{33}{-6}x=\frac{15}{-6}$$

Reduce the fraction \frac{33}{-6} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{11}{2}x=\frac{15}{-6}$$

Reduce the fraction \frac{15}{-6} to lowest terms by extracting and canceling out 3.

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

Divide -\frac{11}{2}, the coefficient of the x term, by 2 to get -\frac{11}{4}. Then add the square of -\frac{11}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{2}x+\left(-\frac{11}{4}\right)^{2}=-\frac{5}{2}+\left(-\frac{11}{4}\right)^{2}$$

Square -\frac{11}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=-\frac{5}{2}+\frac{121}{16}$$

Add -\frac{5}{2} to \frac{121}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{11}{2}x+\frac{121}{16}=\frac{81}{16}$$

Factor x^{2}-\frac{11}{2}x+\frac{121}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{4}\right)^{2}=\frac{81}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{4}\right)^{2}}=\sqrt{\frac{81}{16}}$$

Simplify.

$$x-\frac{11}{4}=\frac{9}{4} x-\frac{11}{4}=-\frac{9}{4}$$

Add \frac{11}{4} to both sides of the equation.

$$x=5 x=\frac{1}{2}$$