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

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) (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
=======================
Cheers,
Stan H.
Answer by Earlsdon(6294) (Show Source): You can put this solution on YOUR website!
Solve for y:
Subtract 3 from both sides.
Factor.
so...
or
or
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