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then for part b by finding the zeros?

and for part c by completing the square?

Yes, use the quadratic formula for part A

ok hang on one sec

Try factoring for part B

Complete the Square for part C

this is what I got for part A what does it mean??

ok factor for part b: (2x+5)(2x−1) which shows the x-intercepts of x=−5/2,1/2

Part C: x=3± 26‾‾‾√ 2

If b^2 - 4ac is negative, then the solutions for the equation for part A are not real.

ok thats what I did but i got that answer, so woulf that mean that it has no solutions??

so are they negative?

Compute b^2 - 4ac

there is not solutions because when you plug in 16^2 -4(9)(60) you get 32 - 2160 = -2128

Well, whatever the number is, it IS negative, but the number is not correct.

I believe that's the intent, yes.

ok thank you and for part b and c factor one and complete the square for the other?

Yes, correct.

but what would i put to say why i choose it?

Post it as a separate question.

thanks so much! @TheSmartOne