SOLUTION: Sorry I just wanted to make sure the problem I sumitted earlier was correct. The correct problem is:2y^2+5y=3 read as follows:2y(squared)+5y=3 did not understand where the

Algebra ->  Quadratic Equations and Parabolas  -> Quadratic Equations Lessons  -> Quadratic Equation Lesson -> SOLUTION: Sorry I just wanted to make sure the problem I sumitted earlier was correct. The correct problem is:2y^2+5y=3 read as follows:2y(squared)+5y=3 did not understand where the      Log On


   



Question 263044: Sorry I just wanted to make sure the problem I sumitted earlier was correct.
The correct problem is:2y^2+5y=3
read as follows:2y(squared)+5y=3
did not understand where the x is coming from in the previous solution, there is no x in the problem.

Found 2 solutions by stanbon, Earlsdon:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
2y^2+5y=3
Rearrange to get:
2y^2 + 5y - 3 = 0
Think of two numbers whose product is 2*-3 = -6
and whose sum = 5
The numbers are +6 and -1
Rewrite your problem replacing the middle term with 6y-y :
2y^2 +6y - y -3 = 0
Factor the 1st two and the last two terms separately:
2y(y+3) -(y+3) = 0
Factor again:
(y+3)(2y-1) = 0
Solve for "y":
y = -3 or y = 1/2
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Cheers,
Stan H.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Solve for y:
2y%5E2%2B5y+=+3 Subtract 3 from both sides.
2y%5E2%2B5y-3+=+0 Factor.
%282y-1%29%28y%2B3%29+=+0 so...
2y+=+1 or y+=+-3
highlight%28y+=+1%2F2%29 or highlight%28y+=+-3%29