Question

use quadratic formula to solve. x^2-3x=-5

Answer 1

0=(1)x^2+(-3)x+(5)
therefore a=1, b=-3, c=5

\[x=(\pm \sqrt{b^{2}-4ac})/2a\]

Answer 2

so x=(sqrt(-11))/2

if your answers are supposed to be real (likely unless you know what complex numbers are) then there is no solution

Answer 3

thats what I got too but wanted to make sure I didnt do something wrong!

Answer 4

that is not simplifted.

Answer 5

simplified*

Answer 6

what isnt simplified?

Answer 7

that answer.

Answer 8

I made the equation x^2-3x+5=0. when I put into the quad equation, I came up with negative 11 under the square root sign. I didn't think you could find the square of a negative number

Answer 9

\[\sqrt{-11}=i \sqrt{11}\]

Answer 10

now i understand what the i in the explanation of how to answer the question. so is the answer \[-3\pm i \sqrt{-11}/2\]

Answer 11

\[\frac{3 \pm i \sqrt {11}}{2}\]

Answer 12

thats what i meant to type! lol

Answer 13

haha, yes i knew..