How do you solve #5x^2-6x-4=0# using the quadratic formula?

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Answer 1 · 2018-04-13T11:35:00

The roots are #(3+sqrt29)/5# or #(3-sqrt29)/5#.

#x=(-b±sqrt(b^2-4ac))/(2a)#

#x=6±sqrt((-6)^2-4*5*(-4))/(2*5)#

#x=(6±sqrt(36+80))/(10)#

#x=(6±sqrt(116))/(10)#

#x=(6±2sqrt(29))/(10)#

#x=(3±sqrt(29))/(5)#

Therefore the roots can be #(3+sqrt29)/5# or #(3-sqrt29)/5#.

Answer 2 · 2018-04-13T11:44:05

Solution: #x= 3/5 +- sqrt 29 /5#

# 5 x^2- 6 x -4 =0 #

Comparing with standard quadratic equation #ax^2+bx+c=0#

# a= 5 ,b=-6 ,c=- 4#. Discriminant, # D= b^2-4 a c# or

#D=36+80 =116#, discriminant is positive, we get two real

solutions, Quadratic formula: #x= (-b+-sqrtD)/(2a) #or

#x= (6 +- sqrt 116)/10 = 3/5 +- sqrt 29 /5#

Solution: #x= 3/5 +- sqrt 29 /5# [Ans]