How do you solve #(y+5)(y-1)=27#?

1 Answer
Mar 25, 2018

Answer:

#y=4,-8#

Explanation:

multiply the parentheses together
#y^2-y+5y-5=27#

combine like terms
#y^2+4y-5=27#

subtract 27 from both sides
#y^2+4y-32=0#

figure out what numbers add to 4 and -32. It's -4 and 8
#(y-4)(y+8)=0#

for the left side of the equation to equal zero, either y-4 or y+8 must equal zero (because zero times anything makes zero). y-4 is equal to zero when y = 4, and y+8 is equal to zero when y = -8. therefore,
#y=4,-8#