Solve -6x^2+6-2x=x

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

Solve for x

$$x=\frac{\sqrt{17}-1}{4}\approx 0.780776406$$

$$x=\frac{-\sqrt{17}-1}{4}\approx -1.280776406$$

Steps Using the Quadratic Formula

Subtract x from both sides.

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

Combine -2x and -x to get -3x.

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

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute -6 for a, -3 for b, and 6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-3\right)±\sqrt{\left(-3\right)^{2}-4\left(-6\right)\times 6}}{2\left(-6\right)}$$

Square -3.

$$x=\frac{-\left(-3\right)±\sqrt{9-4\left(-6\right)\times 6}}{2\left(-6\right)}$$

Multiply -4 times -6.

$$x=\frac{-\left(-3\right)±\sqrt{9+24\times 6}}{2\left(-6\right)}$$

Multiply 24 times 6.

$$x=\frac{-\left(-3\right)±\sqrt{9+144}}{2\left(-6\right)}$$

Add 9 to 144.

$$x=\frac{-\left(-3\right)±\sqrt{153}}{2\left(-6\right)}$$

Take the square root of 153.

$$x=\frac{-\left(-3\right)±3\sqrt{17}}{2\left(-6\right)}$$

The opposite of -3 is 3.

$$x=\frac{3±3\sqrt{17}}{2\left(-6\right)}$$

Multiply 2 times -6.

$$x=\frac{3±3\sqrt{17}}{-12}$$

Now solve the equation x=\frac{3±3\sqrt{17}}{-12} when ± is plus. Add 3 to 3\sqrt{17}.

$$x=\frac{3\sqrt{17}+3}{-12}$$

Divide 3+3\sqrt{17} by -12.

$$x=\frac{-\sqrt{17}-1}{4}$$

Now solve the equation x=\frac{3±3\sqrt{17}}{-12} when ± is minus. Subtract 3\sqrt{17} from 3.

$$x=\frac{3-3\sqrt{17}}{-12}$$

Divide 3-3\sqrt{17} by -12.

$$x=\frac{\sqrt{17}-1}{4}$$

The equation is now solved.

$$x=\frac{-\sqrt{17}-1}{4} x=\frac{\sqrt{17}-1}{4}$$

Steps for Completing the Square

Subtract x from both sides.

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

Combine -2x and -x to get -3x.

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

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

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

Divide both sides by -6.

$$\frac{-6x^{2}-3x}{-6}=-\frac{6}{-6}$$

Dividing by -6 undoes the multiplication by -6.

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

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

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

Divide -6 by -6.

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}+\frac{1}{2}x+\left(\frac{1}{4}\right)^{2}=1+\left(\frac{1}{4}\right)^{2}$$

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=1+\frac{1}{16}$$

Add 1 to \frac{1}{16}.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{17}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{17}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{17}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{\sqrt{17}}{4} x+\frac{1}{4}=-\frac{\sqrt{17}}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

$$x=\frac{\sqrt{17}-1}{4} x=\frac{-\sqrt{17}-1}{4}$$