Question 1060525
{{{ f(x) = -2x^2 -5x + 3 }}}
f'(x) = {{{ -4x -5 }}}
f''(x) = {{{ -4 }}}

Since f''(x) < 0, the function is (everywhere) concave down.

Setting f'(x) = 0:  {{{ -4x-5 = 0 }}}  
                            {{{ x = -5/4 }}}

This is a critical point, it is a maximum or minimum.  In this case, since the function is concave down, it is a maximum.  {{{  f(-5/4) = -2(-5/4)^2 - 5(-5/4) + 3 =  (-50/16) + (100/16) + (48/16) = 98/16 }}}

So we know f(x) > 0 at the maximum, and it will be greater than zero for values between the zero crossings of f(x).  To find the zero crossings, set f(x) = 0:

{{{  -2x^2 - 5x + 3 = 0 }}}
{{{    2x^2 + 5x - 3 = 0 }}}
{{{  (2x-1)(x+3) = 0 }}}
{{{  x=1/2 }}} or {{{ x=-3 }}}

So {{{ f(x) > 0 }}}  on the interval {{{ -3 < x < 1/2 }}}