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

Question

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

amistre64 · Answer

let the roots be complex numbers

amistre64 · Answer

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

amistre64 · Answer

i spose that still factors but just not across the reals

cwrw238 · Answer

x^2 - x - 7 = 0

cannot be factored

terenzreignz · Answer

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

cwrw238 · Answer

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

terenzreignz · Answer

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

anonymous · Answer

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

anonymous · Answer

use it...

amistre64 · Answer

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

terenzreignz · Answer

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 

^_^

amistre64 · Answer

yep :)