Question 515573
You are given the quadratic equation

{{{-3x^2 - 7x - 5 = 0}}}

It looks like you are trying to factor this quadratic, but the easiest way to solve this one is to use the quadratic formula: recall that the solutions to

{{{ax^2 + bx + c = 0}}}

are

{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}

Here, a = -3, b = -7, and c = -5, so we substitute in and get:

{{{x = (-(-7) +- sqrt( (-7)^2-4*(-3)*(-5) ))/(2*(-3)) }}}
{{{x = (7 +- sqrt( 49 - 60 ))/(-6) }}} (simplifying)
{{{x = (7 +- sqrt(-11))/(-6) }}} (simplifying)
{{{ x = (7 +- i* sqrt(11))/(-6) }}} (since {{{sqrt(-11) = sqrt(-1)*sqrt(11) = i*sqrt(11)}}})

So the two solutions are:

{{{x = (7 + i * sqrt(11))/(-6) = (-7 - i * sqrt(11))/(6)}}}

and

{{{x = (7 - i * sqrt(11))/(-6) = (-7 + i * sqrt(11))/(6)}}}

Now, the x-intercepts of the graph of {{{f(x) = ax^2 + bx + c}}} are the REAL roots of {{{ax^2 + bx +c = 0}}}.  If the quadratic equation has no real roots, then the graph never intersects the x-axis: it is either always above the x-axis, or always below it.  In this case, the solution to the quadratic equation are non-real complex numbers, so there are no x-intercepts of f(x).