Solve (2x+2)(3-2x)=4x

$$( 2 x + 2 ) ( 3 - 2 x ) = 4 x$$

Solve for x

$$x = -\frac{3}{2} = -1\frac{1}{2} = -1.5$$

$$x=1$$

Steps Using Factoring By Grouping

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

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

Subtract 4x from both sides.

$$2x-4x^{2}+6-4x=0$$

Combine 2x and -4x to get -2x.

$$-2x-4x^{2}+6=0$$

Divide both sides by 2.

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

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

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

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

$$a+b=-1 ab=-2\times 3=-6$$

Since ab is negative, a and b have the opposite signs. Since a+b is negative, the negative number has greater absolute value than the positive. List all such integer pairs that give product -6.

$$1,-6 2,-3$$

Calculate the sum for each pair.

$$1-6=-5 2-3=-1$$

The solution is the pair that gives sum -1.

$$a=2 b=-3$$

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

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

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

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

Factor out common term -x+1 by using distributive property.

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

To find equation solutions, solve -x+1=0 and 2x+3=0.

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

Steps Using the Quadratic Formula

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

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

Subtract 4x from both sides.

$$2x-4x^{2}+6-4x=0$$

Combine 2x and -4x to get -2x.

$$-2x-4x^{2}+6=0$$

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.

$$-4x^{2}-2x+6=0$$

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

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

Square -2.

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

Multiply -4 times -4.

$$x=\frac{-\left(-2\right)±\sqrt{4+16\times 6}}{2\left(-4\right)}$$

Multiply 16 times 6.

$$x=\frac{-\left(-2\right)±\sqrt{4+96}}{2\left(-4\right)}$$

Add 4 to 96.

$$x=\frac{-\left(-2\right)±\sqrt{100}}{2\left(-4\right)}$$

Take the square root of 100.

$$x=\frac{-\left(-2\right)±10}{2\left(-4\right)}$$

The opposite of -2 is 2.

$$x=\frac{2±10}{2\left(-4\right)}$$

Multiply 2 times -4.

$$x=\frac{2±10}{-8}$$

Now solve the equation x=\frac{2±10}{-8} when ± is plus. Add 2 to 10.

$$x=\frac{12}{-8}$$

Reduce the fraction \frac{12}{-8} to lowest terms by extracting and canceling out 4.

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

Now solve the equation x=\frac{2±10}{-8} when ± is minus. Subtract 10 from 2.

$$x=-\frac{8}{-8}$$

Divide -8 by -8.

$$x=1$$

The equation is now solved.

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

Steps for Completing the Square

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

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

Subtract 4x from both sides.

$$2x-4x^{2}+6-4x=0$$

Combine 2x and -4x to get -2x.

$$-2x-4x^{2}+6=0$$

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

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

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.

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

Divide both sides by -4.

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

Dividing by -4 undoes the multiplication by -4.

$$x^{2}+\left(-\frac{2}{-4}\right)x=-\frac{6}{-4}$$

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

$$x^{2}+\frac{1}{2}x=-\frac{6}{-4}$$

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

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

Divide \frac{1}{2}, the coefficient of the x term, by 2 to get \frac{1}{4}. Then add the square of \frac{1}{4} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Square \frac{1}{4} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{3}{2}+\frac{1}{16}$$

Add \frac{3}{2} to \frac{1}{16} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}+\frac{1}{2}x+\frac{1}{16}=\frac{25}{16}$$

Factor x^{2}+\frac{1}{2}x+\frac{1}{16}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x+\frac{1}{4}\right)^{2}=\frac{25}{16}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{4}\right)^{2}}=\sqrt{\frac{25}{16}}$$

Simplify.

$$x+\frac{1}{4}=\frac{5}{4} x+\frac{1}{4}=-\frac{5}{4}$$

Subtract \frac{1}{4} from both sides of the equation.

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