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

1 Answer
Mar 28, 2017

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

Explanation:

First, make sure the equation is in the standard form
ax^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.

2y^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