Question

How do you solve `2x ^ { 2} + 32= - 16x`?

Answer 1

`-4`

Explanation:

`2x^2+32=-16x` <- This is your equation
We will add `16x` to both sides.
That will become
`2x^2+32color(blue)+color(blue)16color(blue)x=-16xcolor(blue)+color(blue)16color(blue)x`

Now simplify it to become
`2x^2+32+16x=0`
Now we will solve this quadratic equation by factoring.
`2x^2+32+16x`
`= 2(x^2+8x+16)`
------->`x^2+8x+16`
Then, we break this equations into groups.
`x^2+8x+16 = (x^2+4x)+(4x+16)` (The `8x` was separated `4x` in each side)

Then we factor out `x` from `x^2+4x` and `4` from `4x+16`.
Which gives us `x(x+4)+4(x+4)`
Then we factor out `(x+4)` to give us `(x+4)(x+4)`.
Which is `2(x+4)(x+4)` and is also `2(x+4)^2`
So, `2(x+4^2)=0`
Now, `x+4=0`.
Now we subtract `4` from both sides.
`x+4color(blue)-color(blue)4=0color(blue)-color(blue)4`
Then we simplify to `x=-4`
So, the answer is `-4`.