Solve 6x^2=x+2.8

$$6 x ^ { 2 } = x + 2.8$$

Solve for x

$$x=\frac{\sqrt{1705}}{60}+\frac{1}{12}\approx 0.771527427$$

$$x=-\frac{\sqrt{1705}}{60}+\frac{1}{12}\approx -0.604860761$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$6x^{2}-x=2.8$$

Subtract 2.8 from both sides.

$$6x^{2}-x-2.8=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -1 for b, and -2.8 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\left(-2.8\right)}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1-24\left(-2.8\right)}}{2\times 6}$$

Multiply -24 times -2.8.

$$x=\frac{-\left(-1\right)±\sqrt{1+67.2}}{2\times 6}$$

Add 1 to 67.2.

$$x=\frac{-\left(-1\right)±\sqrt{68.2}}{2\times 6}$$

Take the square root of 68.2.

$$x=\frac{-\left(-1\right)±\frac{\sqrt{1705}}{5}}{2\times 6}$$

The opposite of -1 is 1.

$$x=\frac{1±\frac{\sqrt{1705}}{5}}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{1±\frac{\sqrt{1705}}{5}}{12}$$

Now solve the equation x=\frac{1±\frac{\sqrt{1705}}{5}}{12} when ± is plus. Add 1 to \frac{\sqrt{1705}}{5}.

$$x=\frac{\frac{\sqrt{1705}}{5}+1}{12}$$

Divide 1+\frac{\sqrt{1705}}{5} by 12.

$$x=\frac{\sqrt{1705}}{60}+\frac{1}{12}$$

Now solve the equation x=\frac{1±\frac{\sqrt{1705}}{5}}{12} when ± is minus. Subtract \frac{\sqrt{1705}}{5} from 1.

$$x=\frac{-\frac{\sqrt{1705}}{5}+1}{12}$$

Divide 1-\frac{\sqrt{1705}}{5} by 12.

$$x=-\frac{\sqrt{1705}}{60}+\frac{1}{12}$$

The equation is now solved.

$$x=\frac{\sqrt{1705}}{60}+\frac{1}{12} x=-\frac{\sqrt{1705}}{60}+\frac{1}{12}$$

Steps for Completing the Square

Subtract x from both sides.

$$6x^{2}-x=2.8$$

Divide both sides by 6.

$$\frac{6x^{2}-x}{6}=\frac{2.8}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{1}{6}x=\frac{2.8}{6}$$

Divide 2.8 by 6.

$$x^{2}-\frac{1}{6}x=\frac{7}{15}$$

Divide -\frac{1}{6}, the coefficient of the x term, by 2 to get -\frac{1}{12}. Then add the square of -\frac{1}{12} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{1}{6}x+\left(-\frac{1}{12}\right)^{2}=\frac{7}{15}+\left(-\frac{1}{12}\right)^{2}$$

Square -\frac{1}{12} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{7}{15}+\frac{1}{144}$$

Add \frac{7}{15} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=\frac{341}{720}$$

Factor x^{2}-\frac{1}{6}x+\frac{1}{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}.

$$\left(x-\frac{1}{12}\right)^{2}=\frac{341}{720}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{12}\right)^{2}}=\sqrt{\frac{341}{720}}$$

Simplify.

$$x-\frac{1}{12}=\frac{\sqrt{1705}}{60} x-\frac{1}{12}=-\frac{\sqrt{1705}}{60}$$

Add \frac{1}{12} to both sides of the equation.

$$x=\frac{\sqrt{1705}}{60}+\frac{1}{12} x=-\frac{\sqrt{1705}}{60}+\frac{1}{12}$$