2y2+5y+2=0

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.

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.

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