SOLUTION: 2 y +13y+40=0 (y+13)=0 (y+ )=0 y+13=0 y=-13

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Question 294611: 2
y +13y+40=0
(y+13)=0 (y+ )=0
y+13=0
y=-13
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!
Solved by pluggable solver: Quadratic Formula
Let's use the quadratic formula to solve for y:


Starting with the general quadratic


ay%5E2%2Bby%2Bc=0


the general solution using the quadratic equation is:


y+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29




So lets solve y%5E2%2B13%2Ay%2B40=0 ( notice a=1, b=13, and c=40)





y+=+%28-13+%2B-+sqrt%28+%2813%29%5E2-4%2A1%2A40+%29%29%2F%282%2A1%29 Plug in a=1, b=13, and c=40




y+=+%28-13+%2B-+sqrt%28+169-4%2A1%2A40+%29%29%2F%282%2A1%29 Square 13 to get 169




y+=+%28-13+%2B-+sqrt%28+169%2B-160+%29%29%2F%282%2A1%29 Multiply -4%2A40%2A1 to get -160




y+=+%28-13+%2B-+sqrt%28+9+%29%29%2F%282%2A1%29 Combine like terms in the radicand (everything under the square root)




y+=+%28-13+%2B-+3%29%2F%282%2A1%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)




y+=+%28-13+%2B-+3%29%2F2 Multiply 2 and 1 to get 2


So now the expression breaks down into two parts


y+=+%28-13+%2B+3%29%2F2 or y+=+%28-13+-+3%29%2F2


Lets look at the first part:


x=%28-13+%2B+3%29%2F2


y=-10%2F2 Add the terms in the numerator

y=-5 Divide


So one answer is

y=-5




Now lets look at the second part:


x=%28-13+-+3%29%2F2


y=-16%2F2 Subtract the terms in the numerator

y=-8 Divide


So another answer is

y=-8


So our solutions are:

y=-5 or y=-8