How do you solve #2x^2 – 3x – 5 = 0#?

1 Answer
Aug 9, 2015

Answer:

#x=-1,5/2#

Explanation:

Factor the equation, note that both 2 and 5 are prime numbers, therefore they can only have themselves and 1 as a factor. Therefore a factorisation of:
#2x^2-3x-5=(2x-a)(x-b)=0#

Is likely.
With either #(absa,absb)=(1,5),(5,1)#

By inspection we see #a=5,b=-1#, therefore we have:
#2x^2-3x-5=(2x-5)(x+1)=0=>x=-1,5/2#