一道初二代数题 多谢了

a^2+2a-14=0
a^2+2a=14

（a-3)(a+6)(a+5)(a-4)+20
=[（a-3)(a+5)][(a+6)(a-4)]+20
=[a^2+2a-15][a^2+2a-24]+20
=[14-15][14-24]+20
=10+20
=30

a平方+2a-14=0
所以有``
a方+2a=14
```````````````
(a-3)(a+6)(a+5)(a-4)+20可转换为
=(a-3)(a+5)(a+6)(a-4)+20
=(a方+2a-15)(a方+2a-24)+20
=(14-15)(14-24)+20
=(-1)(-10)+20
=30

＝(a^2+3a-18)(a^2+a-20)+20
=(a^2+2a-14+a-4)(a^2+2a-14-a-6)+20
=(a-4)(-a-6)+20
=-a^2-2a+24+20
=30

（a-3)(a+6)(a+5)(a-4)+20
（a-3)(a+5)(a+6)(a-4)+20
(a^2+2a-15)(a^2=2a-24)+20
(a^2+2a-14-1)(a^2+2a-14-10)+20
(-1)*(-10)+20
=30