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

Mar 25, 2018

$y = 4 , - 8$

#### Explanation:

multiply the parentheses together
${y}^{2} - y + 5 y - 5 = 27$

combine like terms
${y}^{2} + 4 y - 5 = 27$

subtract 27 from both sides
${y}^{2} + 4 y - 32 = 0$

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

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,
$y = 4 , - 8$