Question

How do you solve `2x^2 – 3x – 5 = 0`?

Answer 1

`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`