Question

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

Answer 1

`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`