How do you find the zeros, real and imaginary, of # y = 75x^2 +100x + 23 # using the quadratic formula?

1 Answer
Jun 4, 2017

Answer:

#x = -2/3 +- (sqrt(31))/15#

Explanation:

Given: #y = 75x^2 + 100x + 23#

Using the quadratic formula requires the equation to be in the form #Ax^2 + Bx +C = 0#

Quadratic formula: #x = (-B +- sqrt(B^2 - 4AC))/(2A)#

For the given, #A = 75, B = 100, C = 23#:

#x = (-100 +- sqrt(10,000 - 4*75*23))/(2*75)#

#x = (-100 +- sqrt(10,000 - 6900))/150 = (-100 +- sqrt(3100))/150#

#x = -10/15 +- sqrt(25*4*31)/150 = -10/15 +- (sqrt(25)sqrt(4)sqrt(31))/150 #

#x = -2/3 +- (10 sqrt(31))/150 = -2/3 +- (sqrt(31))/15#