Question

What is the solution to 2x2 + x + 2 = 0?

Answer 1

@agent0smith

Answer 2

Is that 2 times 2?

Answer 3

its supposed to be 2x squared

Answer 4

\[\large 2x^2+x+2\]

Answer 5

^ like that lol

Answer 6

\[\large \frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]

Answer 7

recognise that formula?

Answer 8

or did they not teach you that yet?

Answer 9

i recognize it i just am not real sure how to complete it once i plug it in

Answer 10

alright, which part of it are you unsure about than?

Answer 11

all of it...im so behind on my work so nothing is making sense because im so overwhelmed.

Answer 12

i know the feeling.

Answer 13

anyway.

the formula above holds try for all equations in this format:

\[ax^2+bx+c=0\]

Answer 14

your equation is exactly like that. so it's pretty much a 1-on-1 copy.

Answer 15

the number in front of the x² is your a, the number in front of your x is your b (if there is no number in front of x you must use 1 instead, if there is no x at all then use 0), the number without an x is your c.

Answer 16

does that make sense?

Answer 17

if that makes sense to you, please tell me which numbers you would pick for a, b and c.

Answer 18

2,1,2?

Answer 19

indeed.

Answer 20

i'll fill them in for you

Answer 21

\[\frac{ -1\pm \sqrt{1^2-4*2*2}}{ 2*2 }\]

Answer 22

solving that will give you the values of x. In this case you can't solve it though, the number under the root is negative.

Answer 23

\[\frac{ -1\sqrt{1^{2}}-16 }{ 4 }\]

Answer 24

thats what i have got so far idk how to do it from here though

Answer 25

did you accidentally miss the 16 under the root or is it supposed to be like that?

Answer 26

if you want to have multiple numbers under a root you can write {} after the root command anything between those two will be under the root.

Answer 27

oh ok, well yes it was supposed to go under it

Answer 28

ye, you can't solve that, means x has no solution. (if you draw a graph you can prove it because x does not actually come close to 0, it doesn't go below 2.