Question 942537
Let {{{ n }}} = the number of years from now  
when the product of their ages is {{{ 378 }}}
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{{{ ( 11 + n )*( 8 + n ) = 378 }}}
{{{ 88 + 8n + 11n + n^2  = 378 }}}
{{{ n^2 + 19n - 290 = 0 }}}
I can use the quadratic formula
{{{ n = ( -b +- sqrt(  b^2 - 4*a*c )) / (2*a) }}}
{{{ a = 1 }}}
{{{ b = 19 }}}
{{{ c = -290 }}}
{{{ n = ( -19 +- sqrt(  19^2 - 4*1*(-290) )) / (2*1) }}}
{{{ n = ( -19 +- sqrt(  361 + 1160 )) / 2 }}}
{{{ n = ( -19 +- sqrt(  1521 )) / 2 }}}
{{{ n = ( -19 + 39) / 2 }}}
{{{ n = 20/2 }}}
{{{ n = 10 }}}
( notice that I can only use the + square root )
In 10 years, the product of their ages will be 378
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check:
{{{ ( 11 + n )*( 8 + n ) = 378 }}}
{{{ ( 11 + 10 )*( 8 + 10 ) = 378 }}}
{{{ 21*18 = 378 }}}
{{{ 378 = 378 }}}
OK