# Solve for y 2y^2+5y+2=0 2y2+5y+2=0
Factor by grouping.
For a polynomial of the form ax2+bx+c, rewrite the middle term as a sum of two terms whose product is a⋅c=2⋅2=4 and whose sum is b=5.
Factor 5 out of 5y.
2y2+5(y)+2=0
Rewrite 5 as 1 plus 4
2y2+(1+4)y+2=0
Apply the distributive property.
2y2+1y+4y+2=0
Multiply y by 1.
2y2+y+4y+2=0
2y2+y+4y+2=0
Factor out the greatest common factor from each group.
Group the first two terms and the last two terms.
(2y2+y)+4y+2=0
Factor out the greatest common factor (GCF) from each group.
y(2y+1)+2(2y+1)=0
y(2y+1)+2(2y+1)=0
Factor the polynomial by factoring out the greatest common factor, 2y+1.
(2y+1)(y+2)=0
(2y+1)(y+2)=0
If any individual factor on the left side of the equation is equal to 0, the entire expression will be equal to 0.
2y+1=0
y+2=0
Set the first factor equal to 0 and solve.
Set the first factor equal to 0.
2y+1=0
Subtract 1 from both sides of the equation.
2y=-1
Divide each term by 2 and simplify.
Divide each term in 2y=-1 by 2.
2y2=-12
Cancel the common factor of 2.
Cancel the common factor.
2y2=-12
Divide y by 1.
y=-12
y=-12
Move the negative in front of the fraction.
y=-12
y=-12
y=-12
Set the next factor equal to 0 and solve.
Set the next factor equal to 0.
y+2=0
Subtract 2 from both sides of the equation.
y=-2
y=-2
The final solution is all the values that make (2y+1)(y+2)=0 true.
y=-12,-2
The result can be shown in multiple forms.
Exact Form:
y=-12,-2
Decimal Form:
y=-0.5,-2
Solve for y 2y^2+5y+2=0

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