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Solve the equation 3y(y+4)=15
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Divide thru by 3 getting: y(y+4)=5 y^2+4y-5=0 factor (y+5)(y-1)= 0 solve y=1 y=-5
first of all multiply the 3y to the brackets so it will be 3y*y=3y^2 ,3y*4=12y and then enter the 15 in the other side : it will be 3y^2+12y-15 and then u can take 3 common : 3 (y^2+4y-5) and then facorize : 3 (y-1) (y+5)
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