How do you factor the expression #36y^2-84y-147#?

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Answer 1 · 2016-03-02T23:03:14

#color(blue)(y=3.49 " or " y= -1.16)# to 2 decimal places

#(y-3.49)" and "(y+1.16)#

Treating this the same way you would as a quadratic #x#

Given:#" "x=36y^2-84y-147#

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

#a=36#
#b=(-84)#
#c=(-146)#

Then:#" " y=(+84+-sqrt((-84)^2-4(36)(-146)))/(2(36))#

#y=(84+-sqrt(7056+21024))/72#

#color(blue)(y=3.49 " or " y= -1.16)# to 2 decimal places

Tony B