Question

4y2(read as 4ysquare)+13y-12 factories

Answer 1

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

Answer 2

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