Solve x*(x-1)=-2*(x^2+x+1)

$$x \cdot ( x - 1 ) = - 2 \cdot ( x ^ { 2 } + x + 1 )$$

Solve for x (complex solution)

$$x=\frac{-1+\sqrt{23}i}{6}\approx -0.166666667+0.799305254i$$

$$x=\frac{-\sqrt{23}i-1}{6}\approx -0.166666667-0.799305254i$$

Steps Using the Quadratic Formula

Use the distributive property to multiply x by x-1.

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

Use the distributive property to multiply -2 by x^{2}+x+1.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

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

Add 2x to both sides.

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

Combine -x and 2x to get x.

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

Add 2 to both sides.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, 1 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-1±\sqrt{1^{2}-4\times 3\times 2}}{2\times 3}$$

Square 1.

$$x=\frac{-1±\sqrt{1-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-1±\sqrt{1-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$x=\frac{-1±\sqrt{1-24}}{2\times 3}$$

Add 1 to -24.

$$x=\frac{-1±\sqrt{-23}}{2\times 3}$$

Take the square root of -23.

$$x=\frac{-1±\sqrt{23}i}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{-1±\sqrt{23}i}{6}$$

Now solve the equation x=\frac{-1±\sqrt{23}i}{6} when ± is plus. Add -1 to i\sqrt{23}.

$$x=\frac{-1+\sqrt{23}i}{6}$$

Now solve the equation x=\frac{-1±\sqrt{23}i}{6} when ± is minus. Subtract i\sqrt{23} from -1.

$$x=\frac{-\sqrt{23}i-1}{6}$$

The equation is now solved.

$$x=\frac{-1+\sqrt{23}i}{6} x=\frac{-\sqrt{23}i-1}{6}$$

Steps for Completing the Square

Use the distributive property to multiply x by x-1.

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

Use the distributive property to multiply -2 by x^{2}+x+1.

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

Add 2x^{2} to both sides.

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

Combine x^{2} and 2x^{2} to get 3x^{2}.

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

Add 2x to both sides.

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

Combine -x and 2x to get x.

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

Divide both sides by 3.

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

Dividing by 3 undoes the multiplication by 3.

$$x^{2}+\frac{1}{3}x=-\frac{2}{3}$$

Divide \frac{1}{3}, the coefficient of the x term, by 2 to get \frac{1}{6}. Then add the square of \frac{1}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=-\frac{2}{3}+\frac{1}{36}$$

Add -\frac{2}{3} to \frac{1}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{3}x+\frac{1}{36}=-\frac{23}{36}$$

Factor x^{2}+\frac{1}{3}x+\frac{1}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{6}\right)^{2}=-\frac{23}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{6}\right)^{2}}=\sqrt{-\frac{23}{36}}$$

Simplify.

$$x+\frac{1}{6}=\frac{\sqrt{23}i}{6} x+\frac{1}{6}=-\frac{\sqrt{23}i}{6}$$

Subtract \frac{1}{6} from both sides of the equation.

$$x=\frac{-1+\sqrt{23}i}{6} x=\frac{-\sqrt{23}i-1}{6}$$