How do you solve 2x^2 – 3x – 5 = 0?

1 Answer
Aug 9, 2015

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