How do you solve $5x + 4x^2 – 4 = 0$ using the quadratic formula?

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Answer 1 · 2015-07-03T03:52:09

Solve y = 4x^2 + 5x - 4 = 0

$D = d^2 = b^2 - 4ac = 25 + 64 = 89 --> d = +-sqrt89$

$x = (-b/(2a)) +- d/(2a) = - 4/10 +- (sqrt89)/10$

How do you solve 5x + 4x^2 – 4 = 0 5x + 4x^2 – 4 = 0 using the quadratic formula?