Solve -x^2+5x/6-1/6=0

$$- x ^ { 2 } + \frac { 5 x } { 6 } - \frac { 1 } { 6 } = 0$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

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

Steps Using Factoring By Grouping

Multiply both sides of the equation by 6.

$$-6x^{2}+5x-1=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -6x^{2}+ax+bx-1. To find a and b, set up a system to be solved.

$$a+b=5 ab=-6\left(-1\right)=6$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 6.

$$1,6 2,3$$

Calculate the sum for each pair.

$$1+6=7 2+3=5$$

The solution is the pair that gives sum 5.

$$a=3 b=2$$

Rewrite -6x^{2}+5x-1 as \left(-6x^{2}+3x\right)+\left(2x-1\right).

$$\left(-6x^{2}+3x\right)+\left(2x-1\right)$$

Factor out -3x in -6x^{2}+3x.

$$-3x\left(2x-1\right)+2x-1$$

Factor out common term 2x-1 by using distributive property.

$$\left(2x-1\right)\left(-3x+1\right)$$

To find equation solutions, solve 2x-1=0 and -3x+1=0.

$$x=\frac{1}{2} x=\frac{1}{3}$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6.

$$-6x^{2}+5x-1=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, 5 for b, and -1 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-5±\sqrt{5^{2}-4\left(-6\right)\left(-1\right)}}{2\left(-6\right)}$$

Square 5.

$$x=\frac{-5±\sqrt{25-4\left(-6\right)\left(-1\right)}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-5±\sqrt{25+24\left(-1\right)}}{2\left(-6\right)}$$

Multiply 24 times -1.

$$x=\frac{-5±\sqrt{25-24}}{2\left(-6\right)}$$

Add 25 to -24.

$$x=\frac{-5±\sqrt{1}}{2\left(-6\right)}$$

Take the square root of 1.

$$x=\frac{-5±1}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{-5±1}{-12}$$

Now solve the equation x=\frac{-5±1}{-12} when ± is plus. Add -5 to 1.

$$x=-\frac{4}{-12}$$

Reduce the fraction \frac{-4}{-12} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{-5±1}{-12} when ± is minus. Subtract 1 from -5.

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

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

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

The equation is now solved.

$$x=\frac{1}{3} x=\frac{1}{2}$$

Steps for Completing the Square

Multiply both sides of the equation by 6.

$$-6x^{2}+5x-1=0$$

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

$$-6x^{2}+5x=1$$

Divide both sides by -6.

$$\frac{-6x^{2}+5x}{-6}=\frac{1}{-6}$$

Dividing by -6 undoes the multiplication by -6.

$$x^{2}+\frac{5}{-6}x=\frac{1}{-6}$$

Divide 5 by -6.

$$x^{2}-\frac{5}{6}x=\frac{1}{-6}$$

Divide 1 by -6.

$$x^{2}-\frac{5}{6}x=-\frac{1}{6}$$

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

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

Square -\frac{5}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{5}{6}x+\frac{25}{144}=-\frac{1}{6}+\frac{25}{144}$$

Add -\frac{1}{6} to \frac{25}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{5}{6}x+\frac{25}{144}=\frac{1}{144}$$

Factor x^{2}-\frac{5}{6}x+\frac{25}{144}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{5}{12}\right)^{2}=\frac{1}{144}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{5}{12} to both sides of the equation.

$$x=\frac{1}{2} x=\frac{1}{3}$$