Solve x-2/4-x^2+2/6x=x+5/3x

$$\frac { x - 2 } { 4 } - \frac { x ^ { 2 } + 2 } { 6 x } = \frac { x + 5 } { 3 x }$$

Solve for x

$$x=-2$$

$$x=12$$

Steps Using Factoring

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12x, the least common multiple of 4,6x,3x.

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

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

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

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

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

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

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

Use the distributive property to multiply 4 by x+5.

$$x^{2}-6x-4=4x+20$$

Subtract 4x from both sides.

$$x^{2}-6x-4-4x=20$$

Combine -6x and -4x to get -10x.

$$x^{2}-10x-4=20$$

Subtract 20 from both sides.

$$x^{2}-10x-4-20=0$$

Subtract 20 from -4 to get -24.

$$x^{2}-10x-24=0$$

To solve the equation, factor x^{2}-10x-24 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.

$$a+b=-10 ab=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$a=-12 b=2$$

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

$$\left(x-12\right)\left(x+2\right)$$

To find equation solutions, solve x-12=0 and x+2=0.

$$x=12 x=-2$$

Steps Using Factoring By Grouping

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12x, the least common multiple of 4,6x,3x.

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

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

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

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

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

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

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

Use the distributive property to multiply 4 by x+5.

$$x^{2}-6x-4=4x+20$$

Subtract 4x from both sides.

$$x^{2}-6x-4-4x=20$$

Combine -6x and -4x to get -10x.

$$x^{2}-10x-4=20$$

Subtract 20 from both sides.

$$x^{2}-10x-4-20=0$$

Subtract 20 from -4 to get -24.

$$x^{2}-10x-24=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-24. To find a and b, set up a system to be solved.

$$a+b=-10 ab=1\left(-24\right)=-24$$

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 -24.

$$1,-24 2,-12 3,-8 4,-6$$

Calculate the sum for each pair.

$$1-24=-23 2-12=-10 3-8=-5 4-6=-2$$

The solution is the pair that gives sum -10.

$$a=-12 b=2$$

Rewrite x^{2}-10x-24 as \left(x^{2}-12x\right)+\left(2x-24\right).

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

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

$$x\left(x-12\right)+2\left(x-12\right)$$

Factor out common term x-12 by using distributive property.

$$\left(x-12\right)\left(x+2\right)$$

To find equation solutions, solve x-12=0 and x+2=0.

$$x=12 x=-2$$

Steps Using the Quadratic Formula

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12x, the least common multiple of 4,6x,3x.

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

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

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

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

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

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

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

Use the distributive property to multiply 4 by x+5.

$$x^{2}-6x-4=4x+20$$

Subtract 4x from both sides.

$$x^{2}-6x-4-4x=20$$

Combine -6x and -4x to get -10x.

$$x^{2}-10x-4=20$$

Subtract 20 from both sides.

$$x^{2}-10x-4-20=0$$

Subtract 20 from -4 to get -24.

$$x^{2}-10x-24=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -10 for b, and -24 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-10\right)±\sqrt{\left(-10\right)^{2}-4\left(-24\right)}}{2}$$

Square -10.

$$x=\frac{-\left(-10\right)±\sqrt{100-4\left(-24\right)}}{2}$$

Multiply -4 times -24.

$$x=\frac{-\left(-10\right)±\sqrt{100+96}}{2}$$

Add 100 to 96.

$$x=\frac{-\left(-10\right)±\sqrt{196}}{2}$$

Take the square root of 196.

$$x=\frac{-\left(-10\right)±14}{2}$$

The opposite of -10 is 10.

$$x=\frac{10±14}{2}$$

Now solve the equation x=\frac{10±14}{2} when ± is plus. Add 10 to 14.

$$x=\frac{24}{2}$$

Divide 24 by 2.

$$x=12$$

Now solve the equation x=\frac{10±14}{2} when ± is minus. Subtract 14 from 10.

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

Divide -4 by 2.

$$x=-2$$

The equation is now solved.

$$x=12 x=-2$$

Steps for Completing the Square

Variable x cannot be equal to 0 since division by zero is not defined. Multiply both sides of the equation by 12x, the least common multiple of 4,6x,3x.

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

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

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

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

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

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

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

Use the distributive property to multiply 4 by x+5.

$$x^{2}-6x-4=4x+20$$

Subtract 4x from both sides.

$$x^{2}-6x-4-4x=20$$

Combine -6x and -4x to get -10x.

$$x^{2}-10x-4=20$$

Add 4 to both sides.

$$x^{2}-10x=20+4$$

Add 20 and 4 to get 24.

$$x^{2}-10x=24$$

Divide -10, the coefficient of the x term, by 2 to get -5. Then add the square of -5 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$x^{2}-10x+\left(-5\right)^{2}=24+\left(-5\right)^{2}$$

Square -5.

$$x^{2}-10x+25=24+25$$

Add 24 to 25.

$$x^{2}-10x+25=49$$

Factor x^{2}-10x+25. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-5\right)^{2}=49$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-5\right)^{2}}=\sqrt{49}$$

Simplify.

$$x-5=7 x-5=-7$$

Add 5 to both sides of the equation.

$$x=12 x=-2$$