How do you solve #5x^2-3x-3=0# using the quadratic formula?

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Answer 1 · 2016-05-22T08:30:02

#x=(3+ sqrt(69))/10#

#x=(3- sqrt(69))/10#

The quadratic formula is #(-b± sqrt(b^2-4ac))/(2a)#

In the equation #5x^2-3x-3=0#

#a=5#
#b=-3#
#c=-3#

Therefore,

#(3± sqrt(-3^2-4xx5xx-3))/(2xx5)#

= #(3± sqrt(9--60))/(10)#

#x=(3± sqrt(69))/10#