How do you factor and solve #9x^2 + 7x - 56 = 0#?

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How do you factor and solve #9x^2 + 7x - 56 = 0#?

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Answer 1 · 2016-07-04T03:30:21

(-7 +- sqrt2065)/18

Explanation:

#f(x) = 9x^2 + 7x - 56 = 0#
#D = d^2 = b^2 - 4ac = 49 + 2016 = 2065#
Since D is not a perfect square, this function can't be factored.
Solve f(x) = 0 by the improved quadratic formula.
#D = 2065 # --> #d = +- sqrt(2065)#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = - 7/18 +- sqrt(2065)/18 = (-7 +- sqrt(2065))/18#

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How do you factor and solve #9x^2 + 7x - 56 = 0#?

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