Solve left(3xright)^2-5x-6=0

$${ \left(3x \right) }^{ 2 } -5x-6=0$$

Solve for x

$$x = \frac{\sqrt{241} + 5}{18} \approx 1.140231928$$

$$x=\frac{5-\sqrt{241}}{18}\approx -0.584676372$$

Steps Using the Quadratic Formula

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-5x-6=0$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-5x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 9 for a, -5 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-5\right)±\sqrt{\left(-5\right)^{2}-4\times 9\left(-6\right)}}{2\times 9}$$

Square -5.

$$x=\frac{-\left(-5\right)±\sqrt{25-4\times 9\left(-6\right)}}{2\times 9}$$

Multiply -4 times 9.

$$x=\frac{-\left(-5\right)±\sqrt{25-36\left(-6\right)}}{2\times 9}$$

Multiply -36 times -6.

$$x=\frac{-\left(-5\right)±\sqrt{25+216}}{2\times 9}$$

Add 25 to 216.

$$x=\frac{-\left(-5\right)±\sqrt{241}}{2\times 9}$$

The opposite of -5 is 5.

$$x=\frac{5±\sqrt{241}}{2\times 9}$$

Multiply 2 times 9.

$$x=\frac{5±\sqrt{241}}{18}$$

Now solve the equation x=\frac{5±\sqrt{241}}{18} when ± is plus. Add 5 to \sqrt{241}.

$$x=\frac{\sqrt{241}+5}{18}$$

Now solve the equation x=\frac{5±\sqrt{241}}{18} when ± is minus. Subtract \sqrt{241} from 5.

$$x=\frac{5-\sqrt{241}}{18}$$

The equation is now solved.

$$x=\frac{\sqrt{241}+5}{18} x=\frac{5-\sqrt{241}}{18}$$

Steps for Completing the Square

Expand \left(3x\right)^{2}.

$$3^{2}x^{2}-5x-6=0$$

Calculate 3 to the power of 2 and get 9.

$$9x^{2}-5x-6=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$9x^{2}-5x=6$$

Divide both sides by 9.

$$\frac{9x^{2}-5x}{9}=\frac{6}{9}$$

Dividing by 9 undoes the multiplication by 9.

$$x^{2}-\frac{5}{9}x=\frac{6}{9}$$

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

$$x^{2}-\frac{5}{9}x=\frac{2}{3}$$

Divide -\frac{5}{9}, the coefficient of the x term, by 2 to get -\frac{5}{18}. Then add the square of -\frac{5}{18} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{5}{9}x+\left(-\frac{5}{18}\right)^{2}=\frac{2}{3}+\left(-\frac{5}{18}\right)^{2}$$

Square -\frac{5}{18} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{9}x+\frac{25}{324}=\frac{2}{3}+\frac{25}{324}$$

Add \frac{2}{3} to \frac{25}{324} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{9}x+\frac{25}{324}=\frac{241}{324}$$

Factor x^{2}-\frac{5}{9}x+\frac{25}{324}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{18}\right)^{2}=\frac{241}{324}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{5}{18}\right)^{2}}=\sqrt{\frac{241}{324}}$$

Simplify.

$$x-\frac{5}{18}=\frac{\sqrt{241}}{18} x-\frac{5}{18}=-\frac{\sqrt{241}}{18}$$

Add \frac{5}{18} to both sides of the equation.

$$x=\frac{\sqrt{241}+5}{18} x=\frac{5-\sqrt{241}}{18}$$