解:因为甲得到的算式为$(2x - a)(3x + b)$
$=6x^2+(2b - 3a)x - ab = 6x^2+11x - 10,$
$ $所以$2b - 3a = 11,$$ab = 10,$
$ $乙得到的算式为$(2x + a)(x + b)=2x^2+(2b + a)x $
$+ ab = 2x^2-9x + 10,$
$ $所以$2b + a = -9,$$ab = 10,$
$ $则$\begin {cases}2b - 3a = 11&①\\2b + a = -9&②\end {cases}$
① - ②,得$-4a = 20,$解得$a = -5,$
$ $将$a = -5$代入$①,$得$2b-3×(-5)=11,$
解得$b = -2,$
$ $所以正确的式子为
$(2x - 5)(3x - 2)=6x^2-19x + 10。$