Solve algebra question!!

I'm not sure how to solve
2y(1+y)=4

Divide both sides by 2 to get:

y(1 + y) = 2

Distribute y into 1 + y:

y² + y = 2

Then, convert y² + y = 2 into standard form:

y² + y - 2 = 0

Factor:

(y - 1)(y + 2) = 0

Finally, set each factor to zero and solve for y.

y - 1 = 0 and y + 2 = 0
y = {1,-2}

I hope this helps!

2y(1+y)=4

2y + 2y^2 = 4 put the polynomials in order

2y^2 + 2y = 4

factor to = 0
2y^2 + 2y - 4 = 0

(first, outer, inner, last)
(2y - 2)(y + 2) = 2y^2 + 4y - 2y - 4 which equals 2y^2 + 2y - 4 = 0

2y-2=0
2y=2
y=1

y+2=0
y=-2

{1, -2}

2y(1+y)=4

First, Remove 2 from the equation.

2/2y(1+y) = 4/2

y(1+y) = 2

Then Remove 1 from equation.

y(1-1+y) = 2-1


y =


/check you answer =


2 * 1 ( 1+ 1) =

4!!

2y ( 1+y) = 4
=> 2y + 2y ^2 = 4
=> 2y^2 + 2y - 4 = 0
=> 2y^2 - 2y + 4y - 4 = 0
=> 2y( y - 1) + 4( y - 1) = 0
=> ( y- 1)( 2y + 4) = 0

This is the factoring and
further either y - 1 = 0 or 2y +4 = 0
=> either y = 1 or 2y = - 4
=> either y = 1 or y = - 4/2 = -2

so thats the value of y . Hope u find it easy!!!

You can just distribute so then after it should look like this :
2y+y=4
Then add the y's
3y=4
then divide 4/3
so y=4/3 or 1 1/3

2y(1+y)=4 distribute
2y + 2y^2=4 set it equal to zero to get a quadratic
2y^2+2y-4=0 factor
(2y-2)(y+2) solve for y

y=1,y=-2

Ok.
What's the Question? Idontgetit.
I'm actually good at algebra.

y=-2,1

Y=3