How do you find the roots, real and imaginary, of #y= 5x^2-6x +35 # using the quadratic formula?

1 Answer
marfre
Apr 9, 2017

#x = 3/5 +- (sqrt(166)i)/5#

Explanation:

Use the quadratic formula #x = (-B +- sqrt(B^2 - 4AC))/(2A)# when the quadratic equation is in the form #Ax^2 + Bx +C = 0#

For #y = 5x^2 - 6x + 35 = 0, " "A = 5, B = -6, C = 35#

#x = (-B +- sqrt(B^2 - 4AC))/(2A) = (6 +- sqrt(36 - 4(5)(35)))/(2*5)#

#x = (6 +- sqrt(-664))/10 = 3/5 +- sqrt(-4*166)/10 = 3/5 +- (2 sqrt(166) i)/10#

#x = 3/5 +- (sqrt(166)i)/5#

Related questions
See all questions in Quadratic Formula
Impact of this question
1420 views around the world
You can reuse this answer
Creative Commons License