SOLUTION: Solve. (y+5)(y-1)=27

Question 246150: Solve.
(y+5)(y-1)=27
Found 2 solutions by richwmiller, oberobic:
Answer by richwmiller(17219) (Show Source): You can put this solution on YOUR website!
(y+5)(y-1)=27
expand
y^2+4y-5=27
add 5 to both sides
y^2+4y=32
we will complete the square
take (4/2)^2 which =4
add 4 to both sides
y^2+4y+4=36
factor
(y+2)^2=36
get square root
y+2=+\-6
y+2=6
y+2=-6
subtract 2 from both sides
y=4
y=-8
see below for other ways to do it

Solved by pluggable solver: SOLVE quadratic equation with variable

Quadratic equation (in our case ) has the following solutons:

For these solutions to exist, the discriminant should not be a negative number.

First, we need to compute the discriminant : .

Discriminant d=144 is greater than zero. That means that there are two solutions: .

Quadratic expression can be factored:

Again, the answer is: 4, -8. Here's your graph:

Answer by oberobic(2304) (Show Source): You can put this solution on YOUR website!
Given the setup:

.
We can multiply through:

.
Collecting terms

.
Subtracting 27 from both sides

.
Can we factor 32 into two terms that are 4 apart? 4 * 8 = 32, so we have:

.
So we have to solutions: y = -8 and y = 4.
.
Checking , we have:

Check.
.
Checking for , we have:

Check.
Done.
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