SOLUTION: solve for p 4p^2-12p-91=0 our class is solving these type equations using the quadratic formula: (-b +and- square root of b^2-4ac)/2a so it would help a lot if you showed

Algebra ->  Algebra  -> Quadratic Equations and Parabolas -> SOLUTION: solve for p 4p^2-12p-91=0 our class is solving these type equations using the quadratic formula: (-b +and- square root of b^2-4ac)/2a so it would help a lot if you showed      Log On

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Question 78836: solve for p
4p^2-12p-91=0
our class is solving these type equations using the quadratic formula:
(-b +and- square root of b^2-4ac)/2a
so it would help a lot if you showed how to do it by plugging it into that formula.
i tried the problem once and got no real solution and then plugged it in again and got 6.5,-3.5. and i don't know which one's right because my notes say that the a has to equal 1... but im not sure if thats right or not
Answer by mathdoc314(58) About Me  (Show Source):
You can put this solution on YOUR website!

If you wanted a to be equal to 1 for some reason,
you could divide everything through by a. To use the quadratic formula you do not have to have a=1.
4p^2-12p-91=0
p^2 - 3p - 91/4 = 0
a=4, b=-12, c=-91
p+=+%2812+%2B-+sqrt%28+144%2B16%2A91+%29%29%2F%282%2A4%29+
p+=+%2812+%2B-+sqrt%28+1600+%29%29%2F8+
p+=+%2812+%2B-+40%29%2F8+
p+=+52%2F8+or+-28%2F8+
p+=+6.5+or+-3.5+
You can plug the values into the original equation (or equivalent) to check:
(-3.5)^2 - 3(-3.5) - 91/4 = 49/4 + 42/4 - 91/4 = 0
(6.5)^2 - 3(6.5) - 91/4 = 169/4 - 78/4 - 91/4 = 0