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

1 Answer
Mar 25, 2018

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