SOLUTION: Solve 4y ^2 = 12 I've done this and was wondering if I was on the right path: 4y ^2 = 12 4y^2 - 12 = 0 (2y - 4)( 2y + 3) = 0 2y -4 = 0 add 4 2y = 4 divide by 2 y = 2

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Solve 4y ^2 = 12 I've done this and was wondering if I was on the right path: 4y ^2 = 12 4y^2 - 12 = 0 (2y - 4)( 2y + 3) = 0 2y -4 = 0 add 4 2y = 4 divide by 2 y = 2       Log On


   

Question 311431: Solve 4y ^2 = 12
I've done this and was wondering if I was on the right path:
4y ^2 = 12
4y^2 - 12 = 0
(2y - 4)( 2y + 3) = 0
2y -4 = 0
add 4
2y = 4
divide by 2
y = 2
2y + 3 = 0
subtract 3
2y = -3
y = -3 /2

Answer by JBarnum(2146) About Me  (Show Source):
You can put this solution on YOUR website!
(2y - 4)( 2y + 3) = 0
did u realize what this foils in to? 4y^2-y-12=0 so this would be incorrect
id simply do this:
4y%5E2=12
divide by 4
y%5E2=3
sqrt each side
y=sqrt%283%29
y=-sqrt%283%29