Solve (x-1)(x+2)/12-(x+1)(x-2)/6-1=x-3/3

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

Solve for x

$$x=-2$$

$$x=1$$

Steps Using the Quadratic Formula

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

$$\left(x-1\right)\left(x+2\right)-2\left(x+1\right)\left(x-2\right)-12=4\left(x-3\right)$$

Use the distributive property to multiply x-1 by x+2 and combine like terms.

$$x^{2}+x-2-2\left(x+1\right)\left(x-2\right)-12=4\left(x-3\right)$$

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

$$x^{2}+x-2+\left(-2x-2\right)\left(x-2\right)-12=4\left(x-3\right)$$

Use the distributive property to multiply -2x-2 by x-2 and combine like terms.

$$x^{2}+x-2-2x^{2}+2x+4-12=4\left(x-3\right)$$

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

$$-x^{2}+x-2+2x+4-12=4\left(x-3\right)$$

Combine x and 2x to get 3x.

$$-x^{2}+3x-2+4-12=4\left(x-3\right)$$

Add -2 and 4 to get 2.

$$-x^{2}+3x+2-12=4\left(x-3\right)$$

Subtract 12 from 2 to get -10.

$$-x^{2}+3x-10=4\left(x-3\right)$$

Use the distributive property to multiply 4 by x-3.

$$-x^{2}+3x-10=4x-12$$

Subtract 4x from both sides.

$$-x^{2}+3x-10-4x=-12$$

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

$$-x^{2}-x-10=-12$$

Add 12 to both sides.

$$-x^{2}-x-10+12=0$$

Add -10 and 12 to get 2.

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

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

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

Multiply -4 times -1.

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

Multiply 4 times 2.

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

Add 1 to 8.

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

Take the square root of 9.

$$x=\frac{-\left(-1\right)±3}{2\left(-1\right)}$$

The opposite of -1 is 1.

$$x=\frac{1±3}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{1±3}{-2}$$

Now solve the equation x=\frac{1±3}{-2} when ± is plus. Add 1 to 3.

$$x=\frac{4}{-2}$$

Divide 4 by -2.

$$x=-2$$

Now solve the equation x=\frac{1±3}{-2} when ± is minus. Subtract 3 from 1.

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

Divide -2 by -2.

$$x=1$$

The equation is now solved.

$$x=-2 x=1$$

Steps for Completing the Square

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

$$\left(x-1\right)\left(x+2\right)-2\left(x+1\right)\left(x-2\right)-12=4\left(x-3\right)$$

Use the distributive property to multiply x-1 by x+2 and combine like terms.

$$x^{2}+x-2-2\left(x+1\right)\left(x-2\right)-12=4\left(x-3\right)$$

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

$$x^{2}+x-2+\left(-2x-2\right)\left(x-2\right)-12=4\left(x-3\right)$$

Use the distributive property to multiply -2x-2 by x-2 and combine like terms.

$$x^{2}+x-2-2x^{2}+2x+4-12=4\left(x-3\right)$$

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

$$-x^{2}+x-2+2x+4-12=4\left(x-3\right)$$

Combine x and 2x to get 3x.

$$-x^{2}+3x-2+4-12=4\left(x-3\right)$$

Add -2 and 4 to get 2.

$$-x^{2}+3x+2-12=4\left(x-3\right)$$

Subtract 12 from 2 to get -10.

$$-x^{2}+3x-10=4\left(x-3\right)$$

Use the distributive property to multiply 4 by x-3.

$$-x^{2}+3x-10=4x-12$$

Subtract 4x from both sides.

$$-x^{2}+3x-10-4x=-12$$

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

$$-x^{2}-x-10=-12$$

Add 10 to both sides.

$$-x^{2}-x=-12+10$$

Add -12 and 10 to get -2.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide -1 by -1.

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

Divide -2 by -1.

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

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

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

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

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

Add 2 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{9}{4}$$

Factor x^{2}+x+\frac{1}{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}.

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

Take the square root of both sides of the equation.

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

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=1 x=-2$$