3y^2+7y+4=0 solve quadratic equation
SuryakantaSenapati:
what to find
to find the zeros
Answers
Answered by
1
3y^2+3y+4y+4=0
3y (y+1)+4 (y+1)=0
(y+1)(3y+4)=0
y+1=0
y=-1
3y+4=0
3y=-4
y=-4/3
3y (y+1)+4 (y+1)=0
(y+1)(3y+4)=0
y+1=0
y=-1
3y+4=0
3y=-4
y=-4/3
Answered by
0
3y^2+3y +4y+4
3y(y+1) +4(y+1)
(3y+1)(3y+4)
3y(y+1) +4(y+1)
(3y+1)(3y+4)
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