Solve 3x(3x-2)=48

$$3 x ( 3 x - 2 ) = 48$$

Solve for x

$$x=-2$$

$$x = \frac{8}{3} = 2\frac{2}{3} \approx 2.666666667$$

Steps Using the Quadratic Formula

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

$$9x^{2}-6x=48$$

Subtract 48 from both sides.

$$9x^{2}-6x-48=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -6 for b, and -48 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-6\right)±\sqrt{\left(-6\right)^{2}-4\times 9\left(-48\right)}}{2\times 9}$$

Square -6.

$$x=\frac{-\left(-6\right)±\sqrt{36-4\times 9\left(-48\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-6\right)±\sqrt{36-36\left(-48\right)}}{2\times 9}$$

Multiply -36 times -48.

$$x=\frac{-\left(-6\right)±\sqrt{36+1728}}{2\times 9}$$

Add 36 to 1728.

$$x=\frac{-\left(-6\right)±\sqrt{1764}}{2\times 9}$$

Take the square root of 1764.

$$x=\frac{-\left(-6\right)±42}{2\times 9}$$

The opposite of -6 is 6.

$$x=\frac{6±42}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{6±42}{18}$$

Now solve the equation x=\frac{6±42}{18} when ± is plus. Add 6 to 42.

$$x=\frac{48}{18}$$

Reduce the fraction \frac{48}{18} to lowest terms by extracting and canceling out 6.

$$x=\frac{8}{3}$$

Now solve the equation x=\frac{6±42}{18} when ± is minus. Subtract 42 from 6.

$$x=-\frac{36}{18}$$

Divide -36 by 18.

$$x=-2$$

The equation is now solved.

$$x=\frac{8}{3} x=-2$$

Steps for Completing the Square

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

$$9x^{2}-6x=48$$

Divide both sides by 9.

$$\frac{9x^{2}-6x}{9}=\frac{48}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}+\left(-\frac{6}{9}\right)x=\frac{48}{9}$$

Reduce the fraction \frac{-6}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{2}{3}x=\frac{48}{9}$$

Reduce the fraction \frac{48}{9} to lowest terms by extracting and canceling out 3.

$$x^{2}-\frac{2}{3}x=\frac{16}{3}$$

Divide -\frac{2}{3}, the coefficient of the x term, by 2 to get -\frac{1}{3}. Then add the square of -\frac{1}{3} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{16}{3}+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{16}{3}+\frac{1}{9}$$

Add \frac{16}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{49}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{49}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{49}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{7}{3} x-\frac{1}{3}=-\frac{7}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{8}{3} x=-2$$