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then the two roots of x are:
\(\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

ya but i got 4+- the square root of -24 / 4 does this make sense?

your b=-5, how did u get 4+/- ... ?

sry the b=-4 and the c=5

You have imaginary roots. Is that what you're worried about?

yup

u actually have imaginary roots for this..

alright can u tell me where to take it from the imaginary roots?

how did u get square root of -24 ?

because with the coefficients, its (-4)^2-4(2)(5) , in the square root right?

lol! b=-5 not -4

no i i mixed up the coefficents in the original question. i was given a=2;b=-4;c=5

ok,
\(\huge{x=\frac{4 \pm \sqrt{-24}}{4}}=\frac{4 \pm 2\sqrt{-6}}{4}=\frac{2\pm i\sqrt{6}}{2}\)

alright thank you very much!