How do you find the zeros, real and imaginary, of y=-5x^2-3x+29 using the quadratic formula?

1 Answer
Jun 12, 2018

(-3 +- sqrt589)/10

Explanation:

y = - 5x^2 - 3x + 29 = 0
Use the improved quadratic formula (Socratic, Google Search).
D = d^2 = b^2 - 4ac = 9 + 580 = 589 --> d = +- sqrt589
There are 2 real roots:
x = -b/(2a) +- d/(2a) = 3/-10 +- sqrt589/10 = (-3 +- sqrt589)/10
x1 = (-3 + sqrt589)/10
x2 = (-3 - sqrt589)/10