Solve x^2+2x-5x-10+8=8x-40+x+2

$$x ^ { 2 } + 2 x - 5 x - 10 + 8 = 8 x - 40 + x + 2$$

Solve for x

$$x=6$$

Steps Using Factoring

Combine 2x and -5x to get -3x.

$$x^{2}-3x-10+8=8x-40+x+2$$

Add -10 and 8 to get -2.

$$x^{2}-3x-2=8x-40+x+2$$

Combine 8x and x to get 9x.

$$x^{2}-3x-2=9x-40+2$$

Add -40 and 2 to get -38.

$$x^{2}-3x-2=9x-38$$

Subtract 9x from both sides.

$$x^{2}-3x-2-9x=-38$$

Combine -3x and -9x to get -12x.

$$x^{2}-12x-2=-38$$

Add 38 to both sides.

$$x^{2}-12x-2+38=0$$

Add -2 and 38 to get 36.

$$x^{2}-12x+36=0$$

To solve the equation, factor x^{2}-12x+36 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.

$$a+b=-12 ab=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -12.

$$a=-6 b=-6$$

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

$$\left(x-6\right)\left(x-6\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x-6=0.

$$x=6$$

Steps Using Factoring By Grouping

Combine 2x and -5x to get -3x.

$$x^{2}-3x-10+8=8x-40+x+2$$

Add -10 and 8 to get -2.

$$x^{2}-3x-2=8x-40+x+2$$

Combine 8x and x to get 9x.

$$x^{2}-3x-2=9x-40+2$$

Add -40 and 2 to get -38.

$$x^{2}-3x-2=9x-38$$

Subtract 9x from both sides.

$$x^{2}-3x-2-9x=-38$$

Combine -3x and -9x to get -12x.

$$x^{2}-12x-2=-38$$

Add 38 to both sides.

$$x^{2}-12x-2+38=0$$

Add -2 and 38 to get 36.

$$x^{2}-12x+36=0$$

To solve the equation, factor the left hand side by grouping. First, left hand side needs to be rewritten as x^{2}+ax+bx+36. To find a and b, set up a system to be solved.

$$a+b=-12 ab=1\times 36=36$$

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 36.

$$-1,-36 -2,-18 -3,-12 -4,-9 -6,-6$$

Calculate the sum for each pair.

$$-1-36=-37 -2-18=-20 -3-12=-15 -4-9=-13 -6-6=-12$$

The solution is the pair that gives sum -12.

$$a=-6 b=-6$$

Rewrite x^{2}-12x+36 as \left(x^{2}-6x\right)+\left(-6x+36\right).

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

Factor out x in the first and -6 in the second group.

$$x\left(x-6\right)-6\left(x-6\right)$$

Factor out common term x-6 by using distributive property.

$$\left(x-6\right)\left(x-6\right)$$

Rewrite as a binomial square.

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

To find equation solution, solve x-6=0.

$$x=6$$

Steps Using the Quadratic Formula

Combine 2x and -5x to get -3x.

$$x^{2}-3x-10+8=8x-40+x+2$$

Add -10 and 8 to get -2.

$$x^{2}-3x-2=8x-40+x+2$$

Combine 8x and x to get 9x.

$$x^{2}-3x-2=9x-40+2$$

Add -40 and 2 to get -38.

$$x^{2}-3x-2=9x-38$$

Subtract 9x from both sides.

$$x^{2}-3x-2-9x=-38$$

Combine -3x and -9x to get -12x.

$$x^{2}-12x-2=-38$$

Add 38 to both sides.

$$x^{2}-12x-2+38=0$$

Add -2 and 38 to get 36.

$$x^{2}-12x+36=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute 1 for a, -12 for b, and 36 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$x=\frac{-\left(-12\right)±\sqrt{\left(-12\right)^{2}-4\times 36}}{2}$$

Square -12.

$$x=\frac{-\left(-12\right)±\sqrt{144-4\times 36}}{2}$$

Multiply -4 times 36.

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

Add 144 to -144.

$$x=\frac{-\left(-12\right)±\sqrt{0}}{2}$$

Take the square root of 0.

$$x=-\frac{-12}{2}$$

The opposite of -12 is 12.

$$x=\frac{12}{2}$$

Divide 12 by 2.

$$x=6$$

Steps for Completing the Square

Combine 2x and -5x to get -3x.

$$x^{2}-3x-10+8=8x-40+x+2$$

Add -10 and 8 to get -2.

$$x^{2}-3x-2=8x-40+x+2$$

Combine 8x and x to get 9x.

$$x^{2}-3x-2=9x-40+2$$

Add -40 and 2 to get -38.

$$x^{2}-3x-2=9x-38$$

Subtract 9x from both sides.

$$x^{2}-3x-2-9x=-38$$

Combine -3x and -9x to get -12x.

$$x^{2}-12x-2=-38$$

Add 2 to both sides.

$$x^{2}-12x=-38+2$$

Add -38 and 2 to get -36.

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

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.

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

Square -6.

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

Add -36 to 36.

$$x^{2}-12x+36=0$$

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}.

$$\left(x-6\right)^{2}=0$$

Take the square root of both sides of the equation.

$$\sqrt{\left(x-6\right)^{2}}=\sqrt{0}$$

Simplify.

$$x-6=0 x-6=0$$

Add 6 to both sides of the equation.

$$x=6 x=6$$

The equation is now solved. Solutions are the same.

$$x=6$$