Solve (2-x)(2x+3)-x(x-2)=0

$$(2-x)(2x+3)-x(x-2)=0$$

Solve for x

$$x=-1$$

$$x=2$$

Steps Using Factoring By Grouping

Use the distributive property to multiply 2-x by 2x+3 and combine like terms.

$$x+6-2x^{2}-x\left(x-2\right)=0$$

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

$$x+6-2x^{2}-\left(x^{2}-2x\right)=0$$

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

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

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

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

Combine x and 2x to get 3x.

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

Divide both sides by 3.

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

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

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

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

$$a+b=1 ab=-2=-2$$

Since ab is negative, a and b have the opposite signs. Since a+b is positive, the positive number has greater absolute value than the negative. The only such pair is the system solution.

$$a=2 b=-1$$

Rewrite -x^{2}+x+2 as \left(-x^{2}+2x\right)+\left(-x+2\right).

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

Factor out -x in the first and -1 in the second group.

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

Factor out common term x-2 by using distributive property.

$$\left(x-2\right)\left(-x-1\right)$$

To find equation solutions, solve x-2=0 and -x-1=0.

$$x=2 x=-1$$

Steps Using the Quadratic Formula

Use the distributive property to multiply 2-x by 2x+3 and combine like terms.

$$x+6-2x^{2}-x\left(x-2\right)=0$$

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

$$x+6-2x^{2}-\left(x^{2}-2x\right)=0$$

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

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

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

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

Combine x and 2x to get 3x.

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

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.

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

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

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

Square 3.

$$x=\frac{-3±\sqrt{9-4\left(-3\right)\times 6}}{2\left(-3\right)}$$

Multiply -4 times -3.

$$x=\frac{-3±\sqrt{9+12\times 6}}{2\left(-3\right)}$$

Multiply 12 times 6.

$$x=\frac{-3±\sqrt{9+72}}{2\left(-3\right)}$$

Add 9 to 72.

$$x=\frac{-3±\sqrt{81}}{2\left(-3\right)}$$

Take the square root of 81.

$$x=\frac{-3±9}{2\left(-3\right)}$$

Multiply 2 times -3.

$$x=\frac{-3±9}{-6}$$

Now solve the equation x=\frac{-3±9}{-6} when ± is plus. Add -3 to 9.

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

Divide 6 by -6.

$$x=-1$$

Now solve the equation x=\frac{-3±9}{-6} when ± is minus. Subtract 9 from -3.

$$x=-\frac{12}{-6}$$

Divide -12 by -6.

$$x=2$$

The equation is now solved.

$$x=-1 x=2$$

Steps for Completing the Square

Use the distributive property to multiply 2-x by 2x+3 and combine like terms.

$$x+6-2x^{2}-x\left(x-2\right)=0$$

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

$$x+6-2x^{2}-\left(x^{2}-2x\right)=0$$

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

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

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

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

Combine x and 2x to get 3x.

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

Subtract 6 from both sides. Anything subtracted from zero gives its negation.

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

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.

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

Divide both sides by -3.

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

Dividing by -3 undoes the multiplication by -3.

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

Divide 3 by -3.

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

Divide -6 by -3.

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

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.

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

Add \frac{1}{2} to both sides of the equation.

$$x=2 x=-1$$