solve for x when x is a real number 2x^2+5x=3

Question

anonymous · Answer

Subtract 3 over to the otherside so we get $$2x ^{2} + 5x - 3 = 0$$

From here do the quadratic formula using the given. This will solve for x.

anonymous · Answer

Using the equation you found ^

acxbox22 · Answer

factor the expression above and you get:
(2x -1 )(x +3 )=0
now,
2x-1=0  and  x+3=0
so...
x=1/2 and -3

anonymous · Answer

-b±√b^2-4ac/2a 
I'm solving it now hold on.

acxbox22 · Answer

you dont have to use quadratic formula if you can factor and solve

anonymous · Answer

Both of our methods work, Acxbox's method is better when you can factor. The quadratic equation is better when you can't factor it.

anonymous · Answer

I just don't see how to factor it out that way.

anonymous · Answer

wait...
dont you multiply the 2 to the 3 in the back and make it 6. so originally it would be 
2x^2+5x-6 and you would factor it then?

anonymous · Answer

No. For the  part where you got -6, all you have to do is multiply the constants together. 
-1 * 3 = -3

anonymous · Answer

wait why -1? why not 1*-3?

anonymous · Answer

even though its the same. The 3 is - not the 1.