How do you solve #7y^2-5y-8=0# using the quadratic formula?

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Answer 1 · 2016-03-16T10:00:32

#y_1=(+5+sqrt(249))/(14)# and #y_2=(+5-sqrt(249))/(14)#

#y=(-b+-sqrt(b^2-4ac))/(2a)#

Let #a=7# and #b=-5# and #c=-8#

#y=(--5+-sqrt((-5)^2-4*7*(-8)))/(2*7)#

#y=(+5+-sqrt(25+224))/(14)#

#y=(+5+-sqrt(249))/(14)#

#y_1=(+5+sqrt(249))/(14)#

#y_2=(+5-sqrt(249))/(14)#

God bless.... I hope the explanation is useful.