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

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

Solve for x

$$x=\sqrt{6}+2\approx 4.449489743$$

$$x=2-\sqrt{6}\approx -0.449489743$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 3.

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

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

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

To find the opposite of x^{2}-x-2, find the opposite of each term.

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

Add 3 and 2 to get 5.

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

Subtract 3 from both sides.

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

Subtract 3 from 5 to get 2.

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

Add 3x to both sides.

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

Combine x and 3x to get 4x.

$$2-x^{2}+4x=0$$

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.

$$-x^{2}+4x+2=0$$

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

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

Square 4.

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

Multiply -4 times -1.

$$x=\frac{-4±\sqrt{16+4\times 2}}{2\left(-1\right)}$$

Multiply 4 times 2.

$$x=\frac{-4±\sqrt{16+8}}{2\left(-1\right)}$$

Add 16 to 8.

$$x=\frac{-4±\sqrt{24}}{2\left(-1\right)}$$

Take the square root of 24.

$$x=\frac{-4±2\sqrt{6}}{2\left(-1\right)}$$

Multiply 2 times -1.

$$x=\frac{-4±2\sqrt{6}}{-2}$$

Now solve the equation x=\frac{-4±2\sqrt{6}}{-2} when ± is plus. Add -4 to 2\sqrt{6}.

$$x=\frac{2\sqrt{6}-4}{-2}$$

Divide -4+2\sqrt{6} by -2.

$$x=2-\sqrt{6}$$

Now solve the equation x=\frac{-4±2\sqrt{6}}{-2} when ± is minus. Subtract 2\sqrt{6} from -4.

$$x=\frac{-2\sqrt{6}-4}{-2}$$

Divide -4-2\sqrt{6} by -2.

$$x=\sqrt{6}+2$$

The equation is now solved.

$$x=2-\sqrt{6} x=\sqrt{6}+2$$

Steps for Completing the Square

Multiply both sides of the equation by 3.

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

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

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

To find the opposite of x^{2}-x-2, find the opposite of each term.

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

Add 3 and 2 to get 5.

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

Add 3x to both sides.

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

Combine x and 3x to get 4x.

$$5-x^{2}+4x=3$$

Subtract 5 from both sides.

$$-x^{2}+4x=3-5$$

Subtract 5 from 3 to get -2.

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

Divide both sides by -1.

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

Dividing by -1 undoes the multiplication by -1.

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

Divide 4 by -1.

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

Divide -2 by -1.

$$x^{2}-4x=2$$

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.

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

Square -2.

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

Add 2 to 4.

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

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}.

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

Take the square root of both sides of the equation.

$$\sqrt{\left(x-2\right)^{2}}=\sqrt{6}$$

Simplify.

$$x-2=\sqrt{6} x-2=-\sqrt{6}$$

Add 2 to both sides of the equation.

$$x=\sqrt{6}+2 x=2-\sqrt{6}$$