y(square) + 13y + 12?

Answer 1

y = 12 or -1

Answer 2

Basically you have to factor by reversing foil I've pasted a link below to explain:

y^2 + 13y + 12 = (y+12) (y+1) so y = -12 and y = -1,

Answer 3

This is a quadratic formula type of problem. x = [ -b ± sqrt(b^2 - 4ac) ] / 2a
Therefore, let's work this step-by-step.
First let's break the quadratic formula in two steps to be easier for you to understand.
Let's say that @ = b^2 - 4(a)(c)
@ = 13^2 - 4(1)(12)
@ = 169 - 4(12)
@ = 169 -48
@ = 121

Then we do Y = (-b ± square root of @) / 2a
Y = (-13 ± square root of 121) / 2 (1)
Y = (-13 ± 11) / 2
then do Y' = (-13 + 11)/2 = -1
and Y" = (-13 - 11) / 2 = -12

To double check if the answers are correct, insert each one of the Y values that we found in the equation [ y(square) + 13y + 12 ] and see if it is equal to zero.

Answer 4

12 or -1

Answer 5

(y+1)(y+12)

y=-1 ;-12

Answer 6

y^2+13y+12=0

(y+12)(y+1)=0

y=12=0, so y= -12

y+1=0,so y= -1

Answer 7

13y+12=25yy(square)

Answer 8

Use equation:
y(1or2)= -p/2 +or- second root from p(square)/4 -q
where the p is number at y and q is decimal number (in your case is 12). You'll get two solutions (y1 or y2) and therefore you got +or- in equation. Best Regards and veradisca! Neven

Answer 9

this implies...
(y+12)(y+1).
if it is equated to 0,y=-12,-1

Answer 10

(y+12)(y+1) that was kinda easy