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

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Answer 1 · 2017-03-28T22:22:54

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

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#