Solve 2(x-3)=-3x(3-x)

$$2 ( x - 3 ) = - 3 x ( 3 - x )$$

Solve for x

$$x=3$$

$$x=\frac{2}{3}\approx 0.666666667$$

Steps Using Factoring By Grouping

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

$$2x-6=-3x\left(3-x\right)$$

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

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

Add 9x to both sides.

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

Combine 2x and 9x to get 11x.

$$11x-6=3x^{2}$$

Subtract 3x^{2} from both sides.

$$11x-6-3x^{2}=0$$

Rearrange the polynomial to put it in standard form. Place the terms in order from highest to lowest power.

$$-3x^{2}+11x-6=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as -3x^{2}+ax+bx-6. To find a and b, set up a system to be solved.

$$a+b=11 ab=-3\left(-6\right)=18$$

Since ab is positive, a and b have the same sign. Since a+b is positive, a and b are both positive. List all such integer pairs that give product 18.

$$1,18 2,9 3,6$$

Calculate the sum for each pair.

$$1+18=19 2+9=11 3+6=9$$

The solution is the pair that gives sum 11.

$$a=9 b=2$$

Rewrite -3x^{2}+11x-6 as \left(-3x^{2}+9x\right)+\left(2x-6\right).

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

Factor out 3x in the first and -2 in the second group.

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

Factor out common term -x+3 by using distributive property.

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

To find equation solutions, solve -x+3=0 and 3x-2=0.

$$x=3 x=\frac{2}{3}$$

Steps Using the Quadratic Formula

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

$$2x-6=-3x\left(3-x\right)$$

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

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

Add 9x to both sides.

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

Combine 2x and 9x to get 11x.

$$11x-6=3x^{2}$$

Subtract 3x^{2} from both sides.

$$11x-6-3x^{2}=0$$

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.

$$-3x^{2}+11x-6=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -3 for a, 11 for b, and -6 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-11±\sqrt{11^{2}-4\left(-3\right)\left(-6\right)}}{2\left(-3\right)}$$

Square 11.

$$x=\frac{-11±\sqrt{121-4\left(-3\right)\left(-6\right)}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-11±\sqrt{121+12\left(-6\right)}}{2\left(-3\right)}$$

Multiply 12 times -6.

$$x=\frac{-11±\sqrt{121-72}}{2\left(-3\right)}$$

Add 121 to -72.

$$x=\frac{-11±\sqrt{49}}{2\left(-3\right)}$$

Take the square root of 49.

$$x=\frac{-11±7}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-11±7}{-6}$$

Now solve the equation x=\frac{-11±7}{-6} when ± is plus. Add -11 to 7.

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

Reduce the fraction \frac{-4}{-6} to lowest terms by extracting and canceling out 2.

$$x=\frac{2}{3}$$

Now solve the equation x=\frac{-11±7}{-6} when ± is minus. Subtract 7 from -11.

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

Divide -18 by -6.

$$x=3$$

The equation is now solved.

$$x=\frac{2}{3} x=3$$

Steps for Completing the Square

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

$$2x-6=-3x\left(3-x\right)$$

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

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

Add 9x to both sides.

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

Combine 2x and 9x to get 11x.

$$11x-6=3x^{2}$$

Subtract 3x^{2} from both sides.

$$11x-6-3x^{2}=0$$

Add 6 to both sides. Anything plus zero gives itself.

$$11x-3x^{2}=6$$

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.

$$-3x^{2}+11x=6$$

Divide both sides by -3.

$$\frac{-3x^{2}+11x}{-3}=\frac{6}{-3}$$

Dividing by -3 undoes the multiplication by -3.

$$x^{2}+\frac{11}{-3}x=\frac{6}{-3}$$

Divide 11 by -3.

$$x^{2}-\frac{11}{3}x=\frac{6}{-3}$$

Divide 6 by -3.

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

Divide -\frac{11}{3}, the coefficient of the x term, by 2 to get -\frac{11}{6}. Then add the square of -\frac{11}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{11}{3}x+\left(-\frac{11}{6}\right)^{2}=-2+\left(-\frac{11}{6}\right)^{2}$$

Square -\frac{11}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=-2+\frac{121}{36}$$

Add -2 to \frac{121}{36}.

$$x^{2}-\frac{11}{3}x+\frac{121}{36}=\frac{49}{36}$$

Factor x^{2}-\frac{11}{3}x+\frac{121}{36}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-\frac{11}{6}\right)^{2}=\frac{49}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{11}{6}\right)^{2}}=\sqrt{\frac{49}{36}}$$

Simplify.

$$x-\frac{11}{6}=\frac{7}{6} x-\frac{11}{6}=-\frac{7}{6}$$

Add \frac{11}{6} to both sides of the equation.

$$x=3 x=\frac{2}{3}$$