Solve x+1/3+2x(x-3)/6+1/2(x+2)=1-2/6

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

Solve for x (complex solution)

$$x=\frac{1+\sqrt{71}i}{4}\approx 0.25+2.106537443i$$

$$x=\frac{-\sqrt{71}i+1}{4}\approx 0.25-2.106537443i$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 3,6,2.

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

Use the distributive property to multiply 2 by x+1.

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

Use the distributive property to multiply 2x by x-3.

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

Combine 2x and -6x to get -4x.

$$-4x+2+2x^{2}+3\left(x+2\right)=1-2$$

Use the distributive property to multiply 3 by x+2.

$$-4x+2+2x^{2}+3x+6=1-2$$

Combine -4x and 3x to get -x.

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

Add 2 and 6 to get 8.

$$-x+8+2x^{2}=1-2$$

Subtract 2 from 1 to get -1.

$$-x+8+2x^{2}=-1$$

Add 1 to both sides.

$$-x+8+2x^{2}+1=0$$

Add 8 and 1 to get 9.

$$-x+9+2x^{2}=0$$

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.

$$2x^{2}-x+9=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, -1 for b, and 9 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-1\right)±\sqrt{1-4\times 2\times 9}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-\left(-1\right)±\sqrt{1-8\times 9}}{2\times 2}$$

Multiply -8 times 9.

$$x=\frac{-\left(-1\right)±\sqrt{1-72}}{2\times 2}$$

Add 1 to -72.

$$x=\frac{-\left(-1\right)±\sqrt{-71}}{2\times 2}$$

Take the square root of -71.

$$x=\frac{-\left(-1\right)±\sqrt{71}i}{2\times 2}$$

The opposite of -1 is 1.

$$x=\frac{1±\sqrt{71}i}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{1±\sqrt{71}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{71}i}{4} when ± is plus. Add 1 to i\sqrt{71}.

$$x=\frac{1+\sqrt{71}i}{4}$$

Now solve the equation x=\frac{1±\sqrt{71}i}{4} when ± is minus. Subtract i\sqrt{71} from 1.

$$x=\frac{-\sqrt{71}i+1}{4}$$

The equation is now solved.

$$x=\frac{1+\sqrt{71}i}{4} x=\frac{-\sqrt{71}i+1}{4}$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 3,6,2.

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

Use the distributive property to multiply 2 by x+1.

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

Use the distributive property to multiply 2x by x-3.

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

Combine 2x and -6x to get -4x.

$$-4x+2+2x^{2}+3\left(x+2\right)=1-2$$

Use the distributive property to multiply 3 by x+2.

$$-4x+2+2x^{2}+3x+6=1-2$$

Combine -4x and 3x to get -x.

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

Add 2 and 6 to get 8.

$$-x+8+2x^{2}=1-2$$

Subtract 2 from 1 to get -1.

$$-x+8+2x^{2}=-1$$

Subtract 8 from both sides.

$$-x+2x^{2}=-1-8$$

Subtract 8 from -1 to get -9.

$$-x+2x^{2}=-9$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$2x^{2}-x=-9$$

Divide both sides by 2.

$$\frac{2x^{2}-x}{2}=-\frac{9}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}-\frac{1}{2}x=-\frac{9}{2}$$

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.

$$x^{2}-\frac{1}{2}x+\left(-\frac{1}{4}\right)^{2}=-\frac{9}{2}+\left(-\frac{1}{4}\right)^{2}$$

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

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{9}{2}+\frac{1}{16}$$

Add -\frac{9}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{1}{2}x+\frac{1}{16}=-\frac{71}{16}$$

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}.

$$\left(x-\frac{1}{4}\right)^{2}=-\frac{71}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{4}\right)^{2}}=\sqrt{-\frac{71}{16}}$$

Simplify.

$$x-\frac{1}{4}=\frac{\sqrt{71}i}{4} x-\frac{1}{4}=-\frac{\sqrt{71}i}{4}$$

Add \frac{1}{4} to both sides of the equation.

$$x=\frac{1+\sqrt{71}i}{4} x=\frac{-\sqrt{71}i+1}{4}$$