Solve 0=5x^2-12x-1

$$0 = 5 x ^ { 2 } - 12 x - 1$$

Solve for x

$$x = \frac{\sqrt{41} + 6}{5} \approx 2.480624847$$

$$x=\frac{6-\sqrt{41}}{5}\approx -0.080624847$$

Steps Using the Quadratic Formula

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

$$5x^{2}-12x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 5 for a, -12 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 5\left(-1\right)}}{2\times 5}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 5\left(-1\right)}}{2\times 5}$$

Multiply -4 times 5.

$$x=\frac{-\left(-12\right)±\sqrt{144-20\left(-1\right)}}{2\times 5}$$

Multiply -20 times -1.

$$x=\frac{-\left(-12\right)±\sqrt{144+20}}{2\times 5}$$

Add 144 to 20.

$$x=\frac{-\left(-12\right)±\sqrt{164}}{2\times 5}$$

Take the square root of 164.

$$x=\frac{-\left(-12\right)±2\sqrt{41}}{2\times 5}$$

The opposite of -12 is 12.

$$x=\frac{12±2\sqrt{41}}{2\times 5}$$

Multiply 2 times 5.

$$x=\frac{12±2\sqrt{41}}{10}$$

Now solve the equation x=\frac{12±2\sqrt{41}}{10} when ± is plus. Add 12 to 2\sqrt{41}.

$$x=\frac{2\sqrt{41}+12}{10}$$

Divide 12+2\sqrt{41} by 10.

$$x=\frac{\sqrt{41}+6}{5}$$

Now solve the equation x=\frac{12±2\sqrt{41}}{10} when ± is minus. Subtract 2\sqrt{41} from 12.

$$x=\frac{12-2\sqrt{41}}{10}$$

Divide 12-2\sqrt{41} by 10.

$$x=\frac{6-\sqrt{41}}{5}$$

The equation is now solved.

$$x=\frac{\sqrt{41}+6}{5} x=\frac{6-\sqrt{41}}{5}$$

Steps for Completing the Square

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

$$5x^{2}-12x-1=0$$

Add 1 to both sides. Anything plus zero gives itself.

$$5x^{2}-12x=1$$

Divide both sides by 5.

$$\frac{5x^{2}-12x}{5}=\frac{1}{5}$$

Dividing by 5 undoes the multiplication by 5.

$$x^{2}-\frac{12}{5}x=\frac{1}{5}$$

Divide -\frac{12}{5}, the coefficient of the x term, by 2 to get -\frac{6}{5}. Then add the square of -\frac{6}{5} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{12}{5}x+\left(-\frac{6}{5}\right)^{2}=\frac{1}{5}+\left(-\frac{6}{5}\right)^{2}$$

Square -\frac{6}{5} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=\frac{1}{5}+\frac{36}{25}$$

Add \frac{1}{5} to \frac{36}{25} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{12}{5}x+\frac{36}{25}=\frac{41}{25}$$

Factor x^{2}-\frac{12}{5}x+\frac{36}{25}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{6}{5}\right)^{2}=\frac{41}{25}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{6}{5}\right)^{2}}=\sqrt{\frac{41}{25}}$$

Simplify.

$$x-\frac{6}{5}=\frac{\sqrt{41}}{5} x-\frac{6}{5}=-\frac{\sqrt{41}}{5}$$

Add \frac{6}{5} to both sides of the equation.

$$x=\frac{\sqrt{41}+6}{5} x=\frac{6-\sqrt{41}}{5}$$