Solve 6x^2=x-35

$$6 x ^ { 2 } = x - 35$$

Solve for x (complex solution)

$$x=\frac{1+\sqrt{839}i}{12}\approx 0.083333333+2.413791393i$$

$$x=\frac{-\sqrt{839}i+1}{12}\approx 0.083333333-2.413791393i$$

Steps Using the Quadratic Formula

Subtract x from both sides.

$$6x^{2}-x=-35$$

Add 35 to both sides.

$$6x^{2}-x+35=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 6 for a, -1 for b, and 35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 6\times 35}}{2\times 6}$$

Multiply -4 times 6.

$$x=\frac{-\left(-1\right)±\sqrt{1-24\times 35}}{2\times 6}$$

Multiply -24 times 35.

$$x=\frac{-\left(-1\right)±\sqrt{1-840}}{2\times 6}$$

Add 1 to -840.

$$x=\frac{-\left(-1\right)±\sqrt{-839}}{2\times 6}$$

Take the square root of -839.

$$x=\frac{-\left(-1\right)±\sqrt{839}i}{2\times 6}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{839}i}{2\times 6}$$

Multiply 2 times 6.

$$x=\frac{1±\sqrt{839}i}{12}$$

Now solve the equation x=\frac{1±\sqrt{839}i}{12} when ± is plus. Add 1 to i\sqrt{839}.

$$x=\frac{1+\sqrt{839}i}{12}$$

Now solve the equation x=\frac{1±\sqrt{839}i}{12} when ± is minus. Subtract i\sqrt{839} from 1.

$$x=\frac{-\sqrt{839}i+1}{12}$$

The equation is now solved.

$$x=\frac{1+\sqrt{839}i}{12} x=\frac{-\sqrt{839}i+1}{12}$$

Steps for Completing the Square

Subtract x from both sides.

$$6x^{2}-x=-35$$

Divide both sides by 6.

$$\frac{6x^{2}-x}{6}=-\frac{35}{6}$$

Dividing by 6 undoes the multiplication by 6.

$$x^{2}-\frac{1}{6}x=-\frac{35}{6}$$

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.

$$x^{2}-\frac{1}{6}x+\left(-\frac{1}{12}\right)^{2}=-\frac{35}{6}+\left(-\frac{1}{12}\right)^{2}$$

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

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=-\frac{35}{6}+\frac{1}{144}$$

Add -\frac{35}{6} to \frac{1}{144} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{6}x+\frac{1}{144}=-\frac{839}{144}$$

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}.

$$\left(x-\frac{1}{12}\right)^{2}=-\frac{839}{144}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{12}\right)^{2}}=\sqrt{-\frac{839}{144}}$$

Simplify.

$$x-\frac{1}{12}=\frac{\sqrt{839}i}{12} x-\frac{1}{12}=-\frac{\sqrt{839}i}{12}$$

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

$$x=\frac{1+\sqrt{839}i}{12} x=\frac{-\sqrt{839}i+1}{12}$$