Question

How do you find the zeros, real and imaginary, of `y=- 5x^2-2x+10` using the quadratic formula?

Answer 1

`x = (-1 +- sqrt51)/5`

Explanation:

`y = - 5x^2 - 2x + 10 = 0`
Use the new improved quadratic formula (Socratic Search)
`D = d^2 = b^2 - 4ac = 4 + 200 = 204 = 4(51)` --> `d = +- 2sqrt51`
There are 2 real roots:
`x = -b/(2a) +- d/(2a) = 2/-10 +- (2sqrt51)/-10 = -1/5 +- sqrt51/5 =`
`x = (-1 +- sqrt51)/5`