阅读理解
解方程时,我们经常将整体多次出现的部分打包进行换元处理,从而达到了降次、转整等目的,这一"神奇"的方法叫换元法.
例如:解方程:$(x^{2}-x)^{2}-8(x^{2}-x)+12=0$.
解:设$x^{2}-x=y$.原方程化为$y^{2}-8y+12=0.\therefore \left(y-2\right)\left(y-6\right)=0.\therefore y-2=0$或$y-6=0.\therefore y_{1}=2$,$y_{2}=6$.
当$y=2$时,即$x^{2}-x=2.\therefore \left(x-2\right)\left(x+1\right)=0$,$\therefore x-2=0$或$x+1=0.\therefore x_{1}=2$,$x_{2}=-1$
当$y=6$时,即$x^{2}-x=6.\therefore \left(x-3\right)\left(x+2\right)=0.\therefore x-3=0$或$x+2=0.\therefore x_{3}=3$,$x_{4}=-2.\therefore$原方程的解是$x_{1}=2$,$x_{2}=-1$,$x_{3}=3$,$x_{4}=-2$.
请你利用换元法解方程:$(x^{2}-7)^{2}-(x^{2}-7)-2=0$.