By factoring, x equals both -5 and 7.
Answer: ![(-\infty, -4]](https://tex.z-dn.net/?f=%28-%5Cinfty%2C%20-4%5D)
Curved parenthesis at negative infinity
Square bracket at -4
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Work Shown:

The last inequality shown above is the same as saying 
Converting this to interval notation leads to the final answer of ![(-\infty , -4]](https://tex.z-dn.net/?f=%28-%5Cinfty%20%2C%20-4%5D)
Note the use of a square bracket at -4 to include this endpoint. We can never include either infinity, so we always use a parenthesis for either infinity.
Ok, first put in the -2 for each b. That gives:
|-4(-2)-8|+|-1(-(-2))^2|+2(-2)^3
Let's do each section.
The first section is |-4(-2)-8)|
-4 times -2 is 8, minus 8 is 0. The absolute value of 0 is still 0.
Now we move on to |-1(-(-2))^2)|
First we do exponents
-(-2) is 2, and 2^2 is 4. 4 times -1 is -4. The absolute value of -4 is 4
Now the last section, 2(-2)^3
Exponents first: (-2)^3 is -2 * -2 * -2, which is -8.
-8*2=-16.
0+4+(-16)=-12
Answer:
x=8 & x=3
Step-by-step explanation:
We want to get the (x)s all on one side.
add 5x to each side

Since x^2 is x•x and 5x is 5•x then we can change the formula to
x(x+5)=8
then seperate that
x=8
x+5=8
subtract 5 from each side
x=3.
The solutions are 8 and 3.