Question

How do you factor and solve `2y^2 +7y + 3 =0`?

Answer 1

Factor is `(2y+1)(y+3)`
Solving; `y=-1/2` and `y=-3`

Explanation:

First, make sure the equation is in the standard form
a`x^2` + bx + c = 0

`2y^2 + 7y +3 = 0`

If `a!= 1`, the `ac` method is applied.

In this case, `a= 2` so we have to use the `ac` method.

What is the ac method???

The ac method is simply the value of a times the value of c

`a= 2`

`b= 7`

`c=3`

`a*c= 6`

Now we have to find the numbers that will give a product of 6 as well as a sum of 7.

`6 * 1 =6 (ac)`
`6 + 1 = 7 (b)`

Bingo! Now that we have these numbers, we rewrite the equation substituting `6` and `1` for b.

2`y^2` + y + 6y + 3 = 0 ...this is the same as the inital equation when simplified.

Now, we factor by grouping and pulling out the gcf .

`(2y^2 +y) + (6y +3) = 0`

`y(2y+1) + 3(2y+1) = 0`

`(2y+1)(y+3) = 0`

Solve for y
`2y+1=0`
`2y=-1`

Dividing both sides by 2, we get
`(cancel(2)y)/cancel2 = (-1)/2`
`y=-1/2`

Solve the other also for y
`y+3 = 0`
`y=-3`