How do you solve #-12x^2 + 5x +2=0# using the quadratic formula?

Question

4325 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-16T10:09:52

#x_1=-1/4" "#and #x_2=2/3#

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

Let #a=-12# and #b=5# and #c=2#

#y=(-5+-sqrt((5)^2-4*(-12)*(2)))/(2(-12))#

#y=(-5+-sqrt(25+96))/(-24)#

#y=(-5+-sqrt(121))/(-24)#

#y_1=(-5+11)/(-24)=6/-24=-1/4#

#y_2=(-5-11)/(-24)=-16/(-24)=2/3#

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