Answers

2014-09-23T19:50:17+05:30
By factorization we get
4y²+16y-3y-12=0
take 4 as common in first two terms and -3 as common in next two terms
4y(y+4)-3(y+4)
take y+4 as common
(4y-3)(y+4)=0
by using the principle of zero products
4y-3=0  or  y+4=0
4y=3      or      y=-4
   y=3/4    or    y=-4
thus the values of x are 3/4 and -4
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hey you have some mistake in last line
2014-09-23T19:59:46+05:30
4y^2+13y-12=4y^2+16y-3y-12 = 4y(y+4)-3(y+4) 

The factors are (4y-3)(y+4)

 (4y-3)(y+4) = 0

4y-3=0⇒4y=3⇒y=3/4

y+4=0⇒y=-4

the values of y are 3/4 and -4

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