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breaks down into x+1

and 2x^2+3x+2

2x^2 +3x +1?

+2*

yeah

that only has imaginary roots

i got them

checked by completing the square

is it (-3+7i)/4 and (-3-7i)/4

?

solveing atm

is the question this: (x+1)^2(x-2)+(x+1)(x+2)^2 or this: (x+1)^2(x-2)+(x+1)(x-2)^2

I am getting

-9 +/- i sqrt(27) all over 16

as your roots

wat

23

not 27

squrt 27 = 3sqrt(3)

did i fuck up the quadratic formula?

Should be completing the square for this

why?

Idk, makes more sense to me

are you sure you put the correct formula in wolfram? the thing you wrote initially didn't have complex roots

gimme 15, grabbing lunch

wait no

im finishing this now

so i have 2x^2+3x+2

so -3+sqrt(9-16)/4?

right?

+/-

sqrt -5 KappaG

if that is equivelant to the following, yes

unless I did my math wrong

im confuse

nope

not equivalent

wtf am i doing wrong?

completing the square is easier

explain?

wait

I am retarded

trie

forgot to divide the 2 by 2

1 sec

im just confused at this point, like how am i fucking up the quadratic formula

this problem seems too involved for a summer homework

you weren't potato

yes

what you did is correct

My bad

okay

thank fuck

-3?

yes

ok

typo in paint

ok

so much for not using imaginary numbers in calculus

ok

time to eat

You use them a bunch early on

for solving roots, do not do any actual calc with them though

he initially wrote (x+1)^2(x-2)+(x-2)^2)(x+1)=0

I have 4 pages of different math and physics problems between this lad and umbreon NotLikeThis

which is much easier

no complex roots

but the wolfram page changed the second x-2 to x+2

lmao

if he typoed the wolfgram, then that makes my origional guess right

(x + 1) (x - 2) (2 x - 1)

aka what I said it should include

(x+1)(x-2)(x+1+x-2)

(x+1)(x-2)(2x-1)

logs of negative numbers are definitely a thing because of how complex exponents work

it reduces to ln(|x|) + iπ