Question 32008
19x^2+24x+1<0
solve 19x^2+24x+1=0  using quadratic formula..let the 2 roots be a and b in increasing order.
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
{{{x = (-24 +- sqrt( 24^2-4*19*1 ))/(2*19) }}} 
{{{x = (-24 +- sqrt( 576-76 ))/(38) }}}
 {{{x = (-24 +- 10*sqrt(5))/38 )}}}
{{{x = (-24 + 10*sqrt(5))/38 )}}}=b
{{{x = (-24 - 10*sqrt(5))/38 )}}}=a
call the 2 roots a and b 
now then this can be written as (x-a)(x-b)=0
if this product has to be not equal to 0  but less than zero,
then x should lie between a and b.,since then while one factor is positive the other would be negative and the product would be negative.