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

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

Solve for x

$$x=-5$$

$$x=7$$

Steps Using Factoring

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

$$x^{2}+3-2\left(x+4\right)=30$$

Use the distributive property to multiply -2 by x+4.

$$x^{2}+3-2x-8=30$$

Subtract 8 from 3 to get -5.

$$x^{2}-5-2x=30$$

Subtract 30 from both sides.

$$x^{2}-5-2x-30=0$$

Subtract 30 from -5 to get -35.

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

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

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

To solve the equation, factor x^{2}-2x-35 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.

$$a+b=-2 ab=-35$$

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 -35.

$$1,-35 5,-7$$

Calculate the sum for each pair.

$$1-35=-34 5-7=-2$$

The solution is the pair that gives sum -2.

$$a=-7 b=5$$

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

$$\left(x-7\right)\left(x+5\right)$$

To find equation solutions, solve x-7=0 and x+5=0.

$$x=7 x=-5$$

Steps Using Factoring By Grouping

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

$$x^{2}+3-2\left(x+4\right)=30$$

Use the distributive property to multiply -2 by x+4.

$$x^{2}+3-2x-8=30$$

Subtract 8 from 3 to get -5.

$$x^{2}-5-2x=30$$

Subtract 30 from both sides.

$$x^{2}-5-2x-30=0$$

Subtract 30 from -5 to get -35.

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

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

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

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx-35. To find a and b, set up a system to be solved.

$$a+b=-2 ab=1\left(-35\right)=-35$$

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 -35.

$$1,-35 5,-7$$

Calculate the sum for each pair.

$$1-35=-34 5-7=-2$$

The solution is the pair that gives sum -2.

$$a=-7 b=5$$

Rewrite x^{2}-2x-35 as \left(x^{2}-7x\right)+\left(5x-35\right).

$$\left(x^{2}-7x\right)+\left(5x-35\right)$$

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

$$x\left(x-7\right)+5\left(x-7\right)$$

Factor out common term x-7 by using distributive property.

$$\left(x-7\right)\left(x+5\right)$$

To find equation solutions, solve x-7=0 and x+5=0.

$$x=7 x=-5$$

Steps Using the Quadratic Formula

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

$$x^{2}+3-2\left(x+4\right)=30$$

Use the distributive property to multiply -2 by x+4.

$$x^{2}+3-2x-8=30$$

Subtract 8 from 3 to get -5.

$$x^{2}-5-2x=30$$

Subtract 30 from both sides.

$$x^{2}-5-2x-30=0$$

Subtract 30 from -5 to get -35.

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

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.

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

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -2 for b, and -35 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-2\right)±\sqrt{\left(-2\right)^{2}-4\left(-35\right)}}{2}$$

Square -2.

$$x=\frac{-\left(-2\right)±\sqrt{4-4\left(-35\right)}}{2}$$

Multiply -4 times -35.

$$x=\frac{-\left(-2\right)±\sqrt{4+140}}{2}$$

Add 4 to 140.

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

Take the square root of 144.

$$x=\frac{-\left(-2\right)±12}{2}$$

The opposite of -2 is 2.

$$x=\frac{2±12}{2}$$

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

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

Divide 14 by 2.

$$x=7$$

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

$$x=-\frac{10}{2}$$

Divide -10 by 2.

$$x=-5$$

The equation is now solved.

$$x=7 x=-5$$

Steps for Completing the Square

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

$$x^{2}+3-2\left(x+4\right)=30$$

Use the distributive property to multiply -2 by x+4.

$$x^{2}+3-2x-8=30$$

Subtract 8 from 3 to get -5.

$$x^{2}-5-2x=30$$

Add 5 to both sides.

$$x^{2}-2x=30+5$$

Add 30 and 5 to get 35.

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

Divide -2, the coefficient of the x term, by 2 to get -1. Then add the square of -1 to both sides of the equation. This step makes the left hand side of the equation a perfect square.

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

Add 35 to 1.

$$x^{2}-2x+1=36$$

Factor x^{2}-2x+1. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(x-1\right)^{2}=36$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-1\right)^{2}}=\sqrt{36}$$

Simplify.

$$x-1=6 x-1=-6$$

Add 1 to both sides of the equation.

$$x=7 x=-5$$