Solve (x+3)(x-3)-4/2-x-2/3=(x-2)^2+1/6

$$\frac { ( x + 3 ) ( x - 3 ) - 4 } { 2 } - \frac { x - 2 } { 3 } = \frac { ( x - 2 ) ^ { 2 } + 1 } { 6 }$$

Solve for x

$$x=-5$$

$$x=4$$

Steps Using Factoring By Grouping

Multiply both sides of the equation by 6, the least common multiple of 2,3,6.

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

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3\left(x^{2}-9-4\right)-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

Subtract 4 from -9 to get -13.

$$3\left(x^{2}-13\right)-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

Use the distributive property to multiply 3 by x^{2}-13.

$$3x^{2}-39-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

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

$$3x^{2}-39-2x+4=\left(x-2\right)^{2}+1$$

Add -39 and 4 to get -35.

$$3x^{2}-35-2x=\left(x-2\right)^{2}+1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$3x^{2}-35-2x=x^{2}-4x+4+1$$

Add 4 and 1 to get 5.

$$3x^{2}-35-2x=x^{2}-4x+5$$

Subtract x^{2} from both sides.

$$3x^{2}-35-2x-x^{2}=-4x+5$$

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

$$2x^{2}-35-2x=-4x+5$$

Add 4x to both sides.

$$2x^{2}-35-2x+4x=5$$

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

$$2x^{2}-35+2x=5$$

Subtract 5 from both sides.

$$2x^{2}-35+2x-5=0$$

Subtract 5 from -35 to get -40.

$$2x^{2}-40+2x=0$$

Divide both sides by 2.

$$x^{2}-20+x=0$$

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

$$x^{2}+x-20=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-20. To find a and b, set up a system to be solved.

$$a+b=1 ab=1\left(-20\right)=-20$$

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. List all such integer pairs that give product -20.

$$-1,20 -2,10 -4,5$$

Calculate the sum for each pair.

$$-1+20=19 -2+10=8 -4+5=1$$

The solution is the pair that gives sum 1.

$$a=-4 b=5$$

Rewrite x^{2}+x-20 as \left(x^{2}-4x\right)+\left(5x-20\right).

$$\left(x^{2}-4x\right)+\left(5x-20\right)$$

Factor out x in the first and 5 in the second group.

$$x\left(x-4\right)+5\left(x-4\right)$$

Factor out common term x-4 by using distributive property.

$$\left(x-4\right)\left(x+5\right)$$

To find equation solutions, solve x-4=0 and x+5=0.

$$x=4 x=-5$$

Steps Using the Quadratic Formula

Multiply both sides of the equation by 6, the least common multiple of 2,3,6.

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

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3\left(x^{2}-9-4\right)-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

Subtract 4 from -9 to get -13.

$$3\left(x^{2}-13\right)-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

Use the distributive property to multiply 3 by x^{2}-13.

$$3x^{2}-39-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

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

$$3x^{2}-39-2x+4=\left(x-2\right)^{2}+1$$

Add -39 and 4 to get -35.

$$3x^{2}-35-2x=\left(x-2\right)^{2}+1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$3x^{2}-35-2x=x^{2}-4x+4+1$$

Add 4 and 1 to get 5.

$$3x^{2}-35-2x=x^{2}-4x+5$$

Subtract x^{2} from both sides.

$$3x^{2}-35-2x-x^{2}=-4x+5$$

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

$$2x^{2}-35-2x=-4x+5$$

Add 4x to both sides.

$$2x^{2}-35-2x+4x=5$$

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

$$2x^{2}-35+2x=5$$

Subtract 5 from both sides.

$$2x^{2}-35+2x-5=0$$

Subtract 5 from -35 to get -40.

$$2x^{2}-40+2x=0$$

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.

$$2x^{2}+2x-40=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 2 for a, 2 for b, and -40 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-2±\sqrt{2^{2}-4\times 2\left(-40\right)}}{2\times 2}$$

Square 2.

$$x=\frac{-2±\sqrt{4-4\times 2\left(-40\right)}}{2\times 2}$$

Multiply -4 times 2.

$$x=\frac{-2±\sqrt{4-8\left(-40\right)}}{2\times 2}$$

Multiply -8 times -40.

$$x=\frac{-2±\sqrt{4+320}}{2\times 2}$$

Add 4 to 320.

$$x=\frac{-2±\sqrt{324}}{2\times 2}$$

Take the square root of 324.

$$x=\frac{-2±18}{2\times 2}$$

Multiply 2 times 2.

$$x=\frac{-2±18}{4}$$

Now solve the equation x=\frac{-2±18}{4} when ± is plus. Add -2 to 18.

$$x=\frac{16}{4}$$

Divide 16 by 4.

$$x=4$$

Now solve the equation x=\frac{-2±18}{4} when ± is minus. Subtract 18 from -2.

$$x=-\frac{20}{4}$$

Divide -20 by 4.

$$x=-5$$

The equation is now solved.

$$x=4 x=-5$$

Steps for Completing the Square

Multiply both sides of the equation by 6, the least common multiple of 2,3,6.

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

Consider \left(x+3\right)\left(x-3\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}. Square 3.

$$3\left(x^{2}-9-4\right)-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

Subtract 4 from -9 to get -13.

$$3\left(x^{2}-13\right)-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

Use the distributive property to multiply 3 by x^{2}-13.

$$3x^{2}-39-2\left(x-2\right)=\left(x-2\right)^{2}+1$$

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

$$3x^{2}-39-2x+4=\left(x-2\right)^{2}+1$$

Add -39 and 4 to get -35.

$$3x^{2}-35-2x=\left(x-2\right)^{2}+1$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(x-2\right)^{2}.

$$3x^{2}-35-2x=x^{2}-4x+4+1$$

Add 4 and 1 to get 5.

$$3x^{2}-35-2x=x^{2}-4x+5$$

Subtract x^{2} from both sides.

$$3x^{2}-35-2x-x^{2}=-4x+5$$

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

$$2x^{2}-35-2x=-4x+5$$

Add 4x to both sides.

$$2x^{2}-35-2x+4x=5$$

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

$$2x^{2}-35+2x=5$$

Add 35 to both sides.

$$2x^{2}+2x=5+35$$

Add 5 and 35 to get 40.

$$2x^{2}+2x=40$$

Divide both sides by 2.

$$\frac{2x^{2}+2x}{2}=\frac{40}{2}$$

Dividing by 2 undoes the multiplication by 2.

$$x^{2}+\frac{2}{2}x=\frac{40}{2}$$

Divide 2 by 2.

$$x^{2}+x=\frac{40}{2}$$

Divide 40 by 2.

$$x^{2}+x=20$$

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.

$$x^{2}+x+\left(\frac{1}{2}\right)^{2}=20+\left(\frac{1}{2}\right)^{2}$$

Square \frac{1}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}+x+\frac{1}{4}=20+\frac{1}{4}$$

Add 20 to \frac{1}{4}.

$$x^{2}+x+\frac{1}{4}=\frac{81}{4}$$

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}.

$$\left(x+\frac{1}{2}\right)^{2}=\frac{81}{4}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x+\frac{1}{2}\right)^{2}}=\sqrt{\frac{81}{4}}$$

Simplify.

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

Subtract \frac{1}{2} from both sides of the equation.

$$x=4 x=-5$$