Question

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

Answer 1

`(-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`