Solve 0=3x^2-2x-32

$$0 = 3 x ^ { 2 } - 2 x - 32$$

Solve for x

$$x = \frac{\sqrt{97} + 1}{3} \approx 3.616285934$$

$$x=\frac{1-\sqrt{97}}{3}\approx -2.949619267$$

Steps Using the Quadratic Formula

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-2x-32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and -32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-32\right)}}{2\times 3}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-32\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12\left(-32\right)}}{2\times 3}$$

Multiply -12 times -32.

$$x=\frac{-\left(-2\right)±\sqrt{4+384}}{2\times 3}$$

Add 4 to 384.

$$x=\frac{-\left(-2\right)±\sqrt{388}}{2\times 3}$$

Take the square root of 388.

$$x=\frac{-\left(-2\right)±2\sqrt{97}}{2\times 3}$$

The opposite of -2 is 2.

$$x=\frac{2±2\sqrt{97}}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±2\sqrt{97}}{6}$$

Now solve the equation x=\frac{2±2\sqrt{97}}{6} when ± is plus. Add 2 to 2\sqrt{97}.

$$x=\frac{2\sqrt{97}+2}{6}$$

Divide 2+2\sqrt{97} by 6.

$$x=\frac{\sqrt{97}+1}{3}$$

Now solve the equation x=\frac{2±2\sqrt{97}}{6} when ± is minus. Subtract 2\sqrt{97} from 2.

$$x=\frac{2-2\sqrt{97}}{6}$$

Divide 2-2\sqrt{97} by 6.

$$x=\frac{1-\sqrt{97}}{3}$$

The equation is now solved.

$$x=\frac{\sqrt{97}+1}{3} x=\frac{1-\sqrt{97}}{3}$$

Steps for Completing the Square

Swap sides so that all variable terms are on the left hand side.

$$3x^{2}-2x-32=0$$

Add 32 to both sides. Anything plus zero gives itself.

$$3x^{2}-2x=32$$

Divide both sides by 3.

$$\frac{3x^{2}-2x}{3}=\frac{32}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{2}{3}x=\frac{32}{3}$$

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.

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=\frac{32}{3}+\left(-\frac{1}{3}\right)^{2}$$

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

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{32}{3}+\frac{1}{9}$$

Add \frac{32}{3} to \frac{1}{9} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{97}{9}$$

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}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{97}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{97}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{\sqrt{97}}{3} x-\frac{1}{3}=-\frac{\sqrt{97}}{3}$$

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

$$x=\frac{\sqrt{97}+1}{3} x=\frac{1-\sqrt{97}}{3}$$