Question 1123580
Looking at {{{ (3x^2+2x-5)(2x-5) }}}  there will be two multiplications that produce an {{{ x^2}}} term:

{{{ (3x^2)*(-5) }}} + {{{ (2x)*(2x) }}}
= {{{ -15x^2 }}} + {{{ 4x^2 }}} 
=  {{{ -11x^2 }}}

so the coefficient is  {{{ highlight ( -11 ) }}}


You know this because the exponents of x must add to 2  (2+0=2,  1+1=2, and there are no others).  

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The long way would be to multiply out then combine like terms, and then pick the {{{x^2 }}} coefficient:
{{{ (3x^2+2x-5)(2x-5) }}}  
= {{{ (3x^2+2x-5)(2x) + (3x^2+2x-5)(-5) }}}
= {{{ (6x^3 + 4x^2 - 10x) + (-15x^2 - 10x + 25) }}}
= {{{ 6x^3 - 11x^2 - 20x + 25 }}}
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EDITED to make long way answer to be 100% correct (had +5 in trinomial, corrected it to be -5)