Solve (x-2)^2/2+5x+6/6=(x+3)(x-3)/3+6

$$\frac { ( x - 2 ) ^ { 2 } } { 2 } + \frac { 5 x + 6 } { 6 } = \frac { ( x + 3 ) ( x - 3 ) } { 3 } + 6$$

Solve for x

$$x=7$$

$$x=0$$

Steps Using Factoring

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

$$3\left(x-2\right)^{2}+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

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

$$3\left(x^{2}-4x+4\right)+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}-12x+12+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

Combine -12x and 5x to get -7x.

$$3x^{2}-7x+12+6=2\left(x+3\right)\left(x-3\right)+36$$

Add 12 and 6 to get 18.

$$3x^{2}-7x+18=2\left(x+3\right)\left(x-3\right)+36$$

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

$$3x^{2}-7x+18=\left(2x+6\right)\left(x-3\right)+36$$

Use the distributive property to multiply 2x+6 by x-3 and combine like terms.

$$3x^{2}-7x+18=2x^{2}-18+36$$

Add -18 and 36 to get 18.

$$3x^{2}-7x+18=2x^{2}+18$$

Subtract 2x^{2} from both sides.

$$3x^{2}-7x+18-2x^{2}=18$$

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

$$x^{2}-7x+18=18$$

Subtract 18 from both sides.

$$x^{2}-7x+18-18=0$$

Subtract 18 from 18 to get 0.

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

Factor out x.

$$x\left(x-7\right)=0$$

To find equation solutions, solve x=0 and x-7=0.

$$x=0 x=7$$

Steps Using the Quadratic Formula

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

$$3\left(x-2\right)^{2}+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

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

$$3\left(x^{2}-4x+4\right)+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}-12x+12+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

Combine -12x and 5x to get -7x.

$$3x^{2}-7x+12+6=2\left(x+3\right)\left(x-3\right)+36$$

Add 12 and 6 to get 18.

$$3x^{2}-7x+18=2\left(x+3\right)\left(x-3\right)+36$$

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

$$3x^{2}-7x+18=\left(2x+6\right)\left(x-3\right)+36$$

Use the distributive property to multiply 2x+6 by x-3 and combine like terms.

$$3x^{2}-7x+18=2x^{2}-18+36$$

Add -18 and 36 to get 18.

$$3x^{2}-7x+18=2x^{2}+18$$

Subtract 2x^{2} from both sides.

$$3x^{2}-7x+18-2x^{2}=18$$

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

$$x^{2}-7x+18=18$$

Subtract 18 from both sides.

$$x^{2}-7x+18-18=0$$

Subtract 18 from 18 to get 0.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -7 for b, and 0 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

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

Take the square root of \left(-7\right)^{2}.

$$x=\frac{-\left(-7\right)±7}{2}$$

The opposite of -7 is 7.

$$x=\frac{7±7}{2}$$

Now solve the equation x=\frac{7±7}{2} when ± is plus. Add 7 to 7.

$$x=\frac{14}{2}$$

Divide 14 by 2.

$$x=7$$

Now solve the equation x=\frac{7±7}{2} when ± is minus. Subtract 7 from 7.

$$x=\frac{0}{2}$$

Divide 0 by 2.

$$x=0$$

The equation is now solved.

$$x=7 x=0$$

Steps for Completing the Square

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

$$3\left(x-2\right)^{2}+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

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

$$3\left(x^{2}-4x+4\right)+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

Use the distributive property to multiply 3 by x^{2}-4x+4.

$$3x^{2}-12x+12+5x+6=2\left(x+3\right)\left(x-3\right)+36$$

Combine -12x and 5x to get -7x.

$$3x^{2}-7x+12+6=2\left(x+3\right)\left(x-3\right)+36$$

Add 12 and 6 to get 18.

$$3x^{2}-7x+18=2\left(x+3\right)\left(x-3\right)+36$$

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

$$3x^{2}-7x+18=\left(2x+6\right)\left(x-3\right)+36$$

Use the distributive property to multiply 2x+6 by x-3 and combine like terms.

$$3x^{2}-7x+18=2x^{2}-18+36$$

Add -18 and 36 to get 18.

$$3x^{2}-7x+18=2x^{2}+18$$

Subtract 2x^{2} from both sides.

$$3x^{2}-7x+18-2x^{2}=18$$

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

$$x^{2}-7x+18=18$$

Subtract 18 from both sides.

$$x^{2}-7x+18-18=0$$

Subtract 18 from 18 to get 0.

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

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

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

Square -\frac{7}{2} by squaring both the numerator and the denominator of the fraction.

$$x^{2}-7x+\frac{49}{4}=\frac{49}{4}$$

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

$$\left(x-\frac{7}{2}\right)^{2}=\frac{49}{4}$$

Take the square root of both sides of the equation.

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

Simplify.

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

Add \frac{7}{2} to both sides of the equation.

$$x=7 x=0$$