Question 716803
Let the amount that she wants to reduce each 
dimension by =  {{{ x }}}
The new dimensions will be
{{{ 10 - x }}} by {{{ 12 - x }}}
The area of the new image is to be
{{{ ( 10*12 ) / 8 }}}
{{{ 120/8 = ( 10 - x )*( 12 - x ) }}}
{{{ 15 = 120 - 12x - 10x + x^2 }}}
{{{ x^2 -22x + 105 = 0 }}}
Use the quadratic formula
{{{ x = (-b +- sqrt( b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = -22 }}}
{{{ c = 105 }}}
{{{ x = (-(-22) +- sqrt( (-22)^2 - 4*1*105)) / (2*1) }}}
{{{ x = ( 22 +- sqrt( 484 - 420)) / 2 }}}
{{{ x = ( 22 +- sqrt( 64)) / 2 }}}
{{{ x = ( 22 + 8 ) / 2 }}}
{{{ x = 15 }}}
This is bigger than the starting dimensions, so I can't use this result
{{{ x = ( 22 - 8 ) / 2 }}}
{{{ x = 14/2 }}}
{{{ x = 7 }}} This can work
check:
{{{ 120/8 = ( 10 - x )*( 12 - x ) }}}
{{{ 15 = ( 10 - 7 )*( 12 - 7 ) }}}
{{{ 15 = 3*5 }}}
{{{ 15 = 15 }}}
OK
The dimensions of the reduced image are 3 x 5