Question

quadratic equation that has two solutions but cannot be solved by factoring

Answer 1

let the roots be complex numbers

Answer 2

(a+bi - x) (a-bi - x)

Answer 3

i spose that still factors but just not across the reals

Answer 4

x^2 - x - 7 = 0

cannot be factored

Answer 5

ambitious much?
\[\large x^2 + 2 = 0\]
works just fine XD

Answer 6

but has 2 solutions
solve this by either completing the square or use the quadratic fromula

Answer 7

Oops... I meant
\[\large x^2 -2= 0\]
XD
#fail

Answer 8

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

Answer 9

use it...

Answer 10

all quadratics have 2 solutions, they just need not be distinct

Answer 11

Cannot be solved by factoring... :D
LOL technically, even \[\large x^2 - 2 = 0\]
can be factored, right?
haha
\[\large (x+\sqrt2 )(x-\sqrt2) = 0\]

Still though...

meh, nvm

^_^

Answer 12

yep :)