Solve 2x(x-1)+2x=33-x(x-2)

$$2 x ( x - 1 ) + 2 x = 33 - x ( x - 2 )$$

Solve for x

$$x=-3$$

$$x = \frac{11}{3} = 3\frac{2}{3} \approx 3.666666667$$

Steps Using the Quadratic Formula

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

$$2x^{2}-2x+2x=33-x\left(x-2\right)$$

Combine -2x and 2x to get 0.

$$2x^{2}=33-x\left(x-2\right)$$

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

$$2x^{2}=33-\left(x^{2}-2x\right)$$

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

$$2x^{2}=33-x^{2}-\left(-2x\right)$$

The opposite of -2x is 2x.

$$2x^{2}=33-x^{2}+2x$$

Subtract 33 from both sides.

$$2x^{2}-33=-x^{2}+2x$$

Add x^{2} to both sides.

$$2x^{2}-33+x^{2}=2x$$

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

$$3x^{2}-33=2x$$

Subtract 2x from both sides.

$$3x^{2}-33-2x=0$$

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.

$$3x^{2}-2x-33=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -2 for b, and -33 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\times 3\left(-33\right)}}{2\times 3}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\times 3\left(-33\right)}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-2\right)±\sqrt{4-12\left(-33\right)}}{2\times 3}$$

Multiply -12 times -33.

$$x=\frac{-\left(-2\right)±\sqrt{4+396}}{2\times 3}$$

Add 4 to 396.

$$x=\frac{-\left(-2\right)±\sqrt{400}}{2\times 3}$$

Take the square root of 400.

$$x=\frac{-\left(-2\right)±20}{2\times 3}$$

The opposite of -2 is 2.

$$x=\frac{2±20}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{2±20}{6}$$

Now solve the equation x=\frac{2±20}{6} when ± is plus. Add 2 to 20.

$$x=\frac{22}{6}$$

Reduce the fraction \frac{22}{6} to lowest terms by extracting and canceling out 2.

$$x=\frac{11}{3}$$

Now solve the equation x=\frac{2±20}{6} when ± is minus. Subtract 20 from 2.

$$x=-\frac{18}{6}$$

Divide -18 by 6.

$$x=-3$$

The equation is now solved.

$$x=\frac{11}{3} x=-3$$

Steps for Completing the Square

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

$$2x^{2}-2x+2x=33-x\left(x-2\right)$$

Combine -2x and 2x to get 0.

$$2x^{2}=33-x\left(x-2\right)$$

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

$$2x^{2}=33-\left(x^{2}-2x\right)$$

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

$$2x^{2}=33-x^{2}-\left(-2x\right)$$

The opposite of -2x is 2x.

$$2x^{2}=33-x^{2}+2x$$

Add x^{2} to both sides.

$$2x^{2}+x^{2}=33+2x$$

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

$$3x^{2}=33+2x$$

Subtract 2x from both sides.

$$3x^{2}-2x=33$$

Divide both sides by 3.

$$\frac{3x^{2}-2x}{3}=\frac{33}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{2}{3}x=\frac{33}{3}$$

Divide 33 by 3.

$$x^{2}-\frac{2}{3}x=11$$

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

$$x^{2}-\frac{2}{3}x+\left(-\frac{1}{3}\right)^{2}=11+\left(-\frac{1}{3}\right)^{2}$$

Square -\frac{1}{3} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=11+\frac{1}{9}$$

Add 11 to \frac{1}{9}.

$$x^{2}-\frac{2}{3}x+\frac{1}{9}=\frac{100}{9}$$

Factor x^{2}-\frac{2}{3}x+\frac{1}{9}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{1}{3}\right)^{2}=\frac{100}{9}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{1}{3}\right)^{2}}=\sqrt{\frac{100}{9}}$$

Simplify.

$$x-\frac{1}{3}=\frac{10}{3} x-\frac{1}{3}=-\frac{10}{3}$$

Add \frac{1}{3} to both sides of the equation.

$$x=\frac{11}{3} x=-3$$