How do you find the zeros, real and imaginary, of #y=- 5x^2+15x+10 # using the quadratic formula?

1 Answer
Feb 17, 2016

Answer:

The answers are: #x=(3 pm sqrt(17))/2#, which are approximately #-0.56# and #3.56#.

Explanation:

The quadratic formula says the zeros of #y=ax^2+bx+c# are #x=(-b pm sqrt(b^2-4ac))/(2a)#.

For this example, #a=-5#, #b=15#, and #c=10#, so we get:

#x=(-15 pm sqrt( (15)^2-4 * (-5) * 10) )/(2*(-5)) =(15 pm sqrt(225+200))/10#

#=(15 pm sqrt(425))/10=(15 pm sqrt(25)sqrt(17))/10#

#=(15 pm 5sqrt(17))/10=(3 pm sqrt(17))/2#