Solve x+1/3+1/2x-3/4=2x-3/12-frac{x}{6}x

$$\frac { x + 1 } { 3 } + \frac { 1 } { 2 } x - \frac { 3 } { 4 } = \frac { 2 x - 3 } { 12 } - \frac { x } { 6 } x$$

Solve for x

$$x=\sqrt{5}-2\approx 0.236067977$$

$$x=-\left(\sqrt{5}+2\right)\approx -4.236067977$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 12, the least common multiple of 3,2,4,12,6.

$$4\left(x+1\right)+6x-9=2x-3-2xx$$

Multiply x and x to get x^{2}.

$$4\left(x+1\right)+6x-9=2x-3-2x^{2}$$

Use the distributive property to multiply 4 by x+1.

$$4x+4+6x-9=2x-3-2x^{2}$$

Combine 4x and 6x to get 10x.

$$10x+4-9=2x-3-2x^{2}$$

Subtract 9 from 4 to get -5.

$$10x-5=2x-3-2x^{2}$$

Subtract 2x from both sides.

$$10x-5-2x=-3-2x^{2}$$

Combine 10x and -2x to get 8x.

$$8x-5=-3-2x^{2}$$

Subtract -3 from both sides.

$$8x-5-\left(-3\right)=-2x^{2}$$

The opposite of -3 is 3.

$$8x-5+3=-2x^{2}$$

Add 2x^{2} to both sides.

$$8x-5+3+2x^{2}=0$$

Add -5 and 3 to get -2.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 8 for b, and -2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-8±\sqrt{8^{2}-4\times 2\left(-2\right)}}{2\times 2}$$

Square 8.

$$x=\frac{-8±\sqrt{64-4\times 2\left(-2\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-8±\sqrt{64-8\left(-2\right)}}{2\times 2}$$

Multiply -8 times -2.

$$x=\frac{-8±\sqrt{64+16}}{2\times 2}$$

Add 64 to 16.

$$x=\frac{-8±\sqrt{80}}{2\times 2}$$

Take the square root of 80.

$$x=\frac{-8±4\sqrt{5}}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-8±4\sqrt{5}}{4}$$

Now solve the equation x=\frac{-8±4\sqrt{5}}{4} when ± is plus. Add -8 to 4\sqrt{5}.

$$x=\frac{4\sqrt{5}-8}{4}$$

Divide -8+4\sqrt{5} by 4.

$$x=\sqrt{5}-2$$

Now solve the equation x=\frac{-8±4\sqrt{5}}{4} when ± is minus. Subtract 4\sqrt{5} from -8.

$$x=\frac{-4\sqrt{5}-8}{4}$$

Divide -8-4\sqrt{5} by 4.

$$x=-\sqrt{5}-2$$

The equation is now solved.

$$x=\sqrt{5}-2 x=-\sqrt{5}-2$$

Steps for Completing the Square

Multiply both sides of the equation by 12, the least common multiple of 3,2,4,12,6.

$$4\left(x+1\right)+6x-9=2x-3-2xx$$

Multiply x and x to get x^{2}.

$$4\left(x+1\right)+6x-9=2x-3-2x^{2}$$

Use the distributive property to multiply 4 by x+1.

$$4x+4+6x-9=2x-3-2x^{2}$$

Combine 4x and 6x to get 10x.

$$10x+4-9=2x-3-2x^{2}$$

Subtract 9 from 4 to get -5.

$$10x-5=2x-3-2x^{2}$$

Subtract 2x from both sides.

$$10x-5-2x=-3-2x^{2}$$

Combine 10x and -2x to get 8x.

$$8x-5=-3-2x^{2}$$

Add 2x^{2} to both sides.

$$8x-5+2x^{2}=-3$$

Add 5 to both sides.

$$8x+2x^{2}=-3+5$$

Add -3 and 5 to get 2.

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

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.

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

Divide both sides by 2.

$$\frac{2x^{2}+8x}{2}=\frac{2}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{8}{2}x=\frac{2}{2}$$

Divide 8 by 2.

$$x^{2}+4x=\frac{2}{2}$$

Divide 2 by 2.

$$x^{2}+4x=1$$

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

$$x^{2}+4x+2^{2}=1+2^{2}$$

Square 2.

$$x^{2}+4x+4=1+4$$

Add 1 to 4.

$$x^{2}+4x+4=5$$

Factor x^{2}+4x+4. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+2\right)^{2}=5$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+2\right)^{2}}=\sqrt{5}$$

Simplify.

$$x+2=\sqrt{5} x+2=-\sqrt{5}$$

Subtract 2 from both sides of the equation.

$$x=\sqrt{5}-2 x=-\sqrt{5}-2$$