SOLUTION: I'm trying to understand the formula or how to solve these problems. 3y to the 2nd power + y-4=0 2a to the 2nd power =11a-12 3j(j-2)=2(2-j)

Question 360530: I'm trying to understand the formula or how to solve these problems.
3y to the 2nd power + y-4=0
2a to the 2nd power =11a-12
3j(j-2)=2(2-j)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
3y^2 + y - 4=0
Factor:
(3y+4)(y-1) = 0
---
y = [-4/3] or y = 1
==================================

2a^2 = 11a - 12
---
2a^2 - 11a + 12 = 0
2a^2-8a-3a+12 = 0
2a(a-4)-3(a-4) = 0
(a-4)(2a-3) = 0
a = 4 or a = 3/2
=================================
3j(j-2)=2(2-j)
3j(j-2) + 2(j-2) = 0
---
(j-2)(3j+2) = 0
---
j = 2 or j = -2/3
=======================
Cheers,
St
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