factor completely:
y3-12y^2-36y y^3-12y^2-36y
= y(y^2-12y-36)
use the std formula y = -b+-sq rtb^2-4ac/2a where a = 1 b = -12 c = -36
y = -(-12)+-sq rt(-12)^2-4.1(-36)/2.1
= 12+-sq rt144+144/2
= 12+-sq rt2.144/2
= 12+-12.sq rt2/2
= 12(1+-sq rt2)/2
= 6(1+-sq rt2)
therefore the value of y = 0 or value of y = 6(1+-sq rt2)
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