Solve (x^2-4)-(x-2)*(3-2x)=0

$$( { x }^{ 2 } -4)-(x-2) \times (3-2x)=0$$

Solve for x

$$x=\frac{1}{3}\approx 0.333333333$$

$$x=2$$

Steps Using Factoring By Grouping

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

$$x^{2}-4-\left(7x-2x^{2}-6\right)=0$$

To find the opposite of 7x-2x^{2}-6, find the opposite of each term.

$$x^{2}-4-7x+2x^{2}+6=0$$

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

$$3x^{2}-4-7x+6=0$$

Add -4 and 6 to get 2.

$$3x^{2}+2-7x=0$$

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

$$3x^{2}-7x+2=0$$

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

$$a+b=-7 ab=3\times 2=6$$

Since ab is positive, a and b have the same sign. Since a+b is negative, a and b are both negative. List all such integer pairs that give product 6.

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

Calculate the sum for each pair.

$$-1-6=-7 -2-3=-5$$

The solution is the pair that gives sum -7.

$$a=-6 b=-1$$

Rewrite 3x^{2}-7x+2 as \left(3x^{2}-6x\right)+\left(-x+2\right).

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

Factor out 3x in the first and -1 in the second group.

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

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

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

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

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

Steps Using the Quadratic Formula

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

$$x^{2}-4-\left(7x-2x^{2}-6\right)=0$$

To find the opposite of 7x-2x^{2}-6, find the opposite of each term.

$$x^{2}-4-7x+2x^{2}+6=0$$

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

$$3x^{2}-4-7x+6=0$$

Add -4 and 6 to get 2.

$$3x^{2}+2-7x=0$$

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.

$$3x^{2}-7x+2=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 3 for a, -7 for b, and 2 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-7\right)±\sqrt{\left(-7\right)^{2}-4\times 3\times 2}}{2\times 3}$$

Square -7.

$$x=\frac{-\left(-7\right)±\sqrt{49-4\times 3\times 2}}{2\times 3}$$

Multiply -4 times 3.

$$x=\frac{-\left(-7\right)±\sqrt{49-12\times 2}}{2\times 3}$$

Multiply -12 times 2.

$$x=\frac{-\left(-7\right)±\sqrt{49-24}}{2\times 3}$$

Add 49 to -24.

$$x=\frac{-\left(-7\right)±\sqrt{25}}{2\times 3}$$

Take the square root of 25.

$$x=\frac{-\left(-7\right)±5}{2\times 3}$$

The opposite of -7 is 7.

$$x=\frac{7±5}{2\times 3}$$

Multiply 2 times 3.

$$x=\frac{7±5}{6}$$

Now solve the equation x=\frac{7±5}{6} when ± is plus. Add 7 to 5.

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

Divide 12 by 6.

$$x=2$$

Now solve the equation x=\frac{7±5}{6} when ± is minus. Subtract 5 from 7.

$$x=\frac{2}{6}$$

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

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

The equation is now solved.

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

Steps for Completing the Square

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

$$x^{2}-4-\left(7x-2x^{2}-6\right)=0$$

To find the opposite of 7x-2x^{2}-6, find the opposite of each term.

$$x^{2}-4-7x+2x^{2}+6=0$$

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

$$3x^{2}-4-7x+6=0$$

Add -4 and 6 to get 2.

$$3x^{2}+2-7x=0$$

Subtract 2 from both sides. Anything subtracted from zero gives its negation.

$$3x^{2}-7x=-2$$

Divide both sides by 3.

$$\frac{3x^{2}-7x}{3}=-\frac{2}{3}$$

Dividing by 3 undoes the multiplication by 3.

$$x^{2}-\frac{7}{3}x=-\frac{2}{3}$$

Divide -\frac{7}{3}, the coefficient of the x term, by 2 to get -\frac{7}{6}. Then add the square of -\frac{7}{6} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-\frac{7}{3}x+\left(-\frac{7}{6}\right)^{2}=-\frac{2}{3}+\left(-\frac{7}{6}\right)^{2}$$

Square -\frac{7}{6} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=-\frac{2}{3}+\frac{49}{36}$$

Add -\frac{2}{3} to \frac{49}{36} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$x^{2}-\frac{7}{3}x+\frac{49}{36}=\frac{25}{36}$$

Factor x^{2}-\frac{7}{3}x+\frac{49}{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}.

$$\left(x-\frac{7}{6}\right)^{2}=\frac{25}{36}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-\frac{7}{6}\right)^{2}}=\sqrt{\frac{25}{36}}$$

Simplify.

$$x-\frac{7}{6}=\frac{5}{6} x-\frac{7}{6}=-\frac{5}{6}$$

Add \frac{7}{6} to both sides of the equation.

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