Solve left(x-2right)^2/2-x^2+2/3=-4

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

Solve for x

$$x=4$$

$$x=8$$

Steps Using Factoring

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

$$3\left(x-2\right)^{2}-2\left(x^{2}+2\right)=-24$$

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

$$3\left(x^{2}-4x+4\right)-2\left(x^{2}+2\right)=-24$$

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

$$3x^{2}-12x+12-2\left(x^{2}+2\right)=-24$$

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

$$3x^{2}-12x+12-2x^{2}-4=-24$$

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

$$x^{2}-12x+12-4=-24$$

Subtract 4 from 12 to get 8.

$$x^{2}-12x+8=-24$$

Add 24 to both sides.

$$x^{2}-12x+8+24=0$$

Add 8 and 24 to get 32.

$$x^{2}-12x+32=0$$

To solve the equation, factor x^{2}-12x+32 using formula x^{2}+\left(a+b\right)x+ab=\left(x+a\right)\left(x+b\right). To find a and b, set up a system to be solved.

$$a+b=-12 ab=32$$

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 32.

$$-1,-32 -2,-16 -4,-8$$

Calculate the sum for each pair.

$$-1-32=-33 -2-16=-18 -4-8=-12$$

The solution is the pair that gives sum -12.

$$a=-8 b=-4$$

Rewrite factored expression \left(x+a\right)\left(x+b\right) using the obtained values.

$$\left(x-8\right)\left(x-4\right)$$

To find equation solutions, solve x-8=0 and x-4=0.

$$x=8 x=4$$

Steps Using Factoring By Grouping

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

$$3\left(x-2\right)^{2}-2\left(x^{2}+2\right)=-24$$

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

$$3\left(x^{2}-4x+4\right)-2\left(x^{2}+2\right)=-24$$

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

$$3x^{2}-12x+12-2\left(x^{2}+2\right)=-24$$

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

$$3x^{2}-12x+12-2x^{2}-4=-24$$

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

$$x^{2}-12x+12-4=-24$$

Subtract 4 from 12 to get 8.

$$x^{2}-12x+8=-24$$

Add 24 to both sides.

$$x^{2}-12x+8+24=0$$

Add 8 and 24 to get 32.

$$x^{2}-12x+32=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+32. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 32=32$$

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 32.

$$-1,-32 -2,-16 -4,-8$$

Calculate the sum for each pair.

$$-1-32=-33 -2-16=-18 -4-8=-12$$

The solution is the pair that gives sum -12.

$$a=-8 b=-4$$

Rewrite x^{2}-12x+32 as \left(x^{2}-8x\right)+\left(-4x+32\right).

$$\left(x^{2}-8x\right)+\left(-4x+32\right)$$

Factor out x in the first and -4 in the second group.

$$x\left(x-8\right)-4\left(x-8\right)$$

Factor out common term x-8 by using distributive property.

$$\left(x-8\right)\left(x-4\right)$$

To find equation solutions, solve x-8=0 and x-4=0.

$$x=8 x=4$$

Steps Using the Quadratic Formula

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

$$3\left(x-2\right)^{2}-2\left(x^{2}+2\right)=-24$$

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

$$3\left(x^{2}-4x+4\right)-2\left(x^{2}+2\right)=-24$$

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

$$3x^{2}-12x+12-2\left(x^{2}+2\right)=-24$$

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

$$3x^{2}-12x+12-2x^{2}-4=-24$$

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

$$x^{2}-12x+12-4=-24$$

Subtract 4 from 12 to get 8.

$$x^{2}-12x+8=-24$$

Add 24 to both sides.

$$x^{2}-12x+8+24=0$$

Add 8 and 24 to get 32.

$$x^{2}-12x+32=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 32 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 32}}{2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 32}}{2}$$

Multiply -4 times 32.

$$x=\frac{-\left(-12\right)±\sqrt{144-128}}{2}$$

Add 144 to -128.

$$x=\frac{-\left(-12\right)±\sqrt{16}}{2}$$

Take the square root of 16.

$$x=\frac{-\left(-12\right)±4}{2}$$

The opposite of -12 is 12.

$$x=\frac{12±4}{2}$$

Now solve the equation x=\frac{12±4}{2} when ± is plus. Add 12 to 4.

$$x=\frac{16}{2}$$

Divide 16 by 2.

$$x=8$$

Now solve the equation x=\frac{12±4}{2} when ± is minus. Subtract 4 from 12.

$$x=\frac{8}{2}$$

Divide 8 by 2.

$$x=4$$

The equation is now solved.

$$x=8 x=4$$

Steps for Completing the Square

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

$$3\left(x-2\right)^{2}-2\left(x^{2}+2\right)=-24$$

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

$$3\left(x^{2}-4x+4\right)-2\left(x^{2}+2\right)=-24$$

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

$$3x^{2}-12x+12-2\left(x^{2}+2\right)=-24$$

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

$$3x^{2}-12x+12-2x^{2}-4=-24$$

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

$$x^{2}-12x+12-4=-24$$

Subtract 4 from 12 to get 8.

$$x^{2}-12x+8=-24$$

Subtract 8 from both sides.

$$x^{2}-12x=-24-8$$

Subtract 8 from -24 to get -32.

$$x^{2}-12x=-32$$

Divide -12, the coefficient of the x term, by 2 to get -6. Then add the square of -6 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-12x+\left(-6\right)^{2}=-32+\left(-6\right)^{2}$$

Square -6.

$$x^{2}-12x+36=-32+36$$

Add -32 to 36.

$$x^{2}-12x+36=4$$

Factor x^{2}-12x+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}.

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

Take the square root of both sides of the equation.

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

Simplify.

$$x-6=2 x-6=-2$$

Add 6 to both sides of the equation.

$$x=8 x=4$$