Answer:
![x_(1) =-6\\x_(2) =1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t489zqc6xp4fnn71gtk2wo6izp68fbh9i0.png)
Explanation:
The given equation is
![(x+2)(x+3)=12](https://img.qammunity.org/2020/formulas/mathematics/middle-school/nr3uj7d5b4uzh92eidh8qfw5fm7jadb2kz.png)
First, we need to apply distributive property to have the quadratic expression
![x^(2) +3x+2x+6=12\\x^(2) +5x+6-12=0\\x^(2) +5x-6=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/m3tsd2vyjui4eaxq6hxycppphbdzy4fqiv.png)
We need to find two numbers which product is -6 and which difference is 5. Those numbers are 6 and 1, because 6 - 1 = 5, and 6(-1) = -6.
![x^(2) +5x-6=(x+6)(x-1)=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/dmqywwvbupukxyr8yjkzf1boztwdf04myh.png)
If we apply the zero product property, that means each binomial is equal to zero
![x+6=0\\x-1=0](https://img.qammunity.org/2020/formulas/mathematics/middle-school/lwns3zzwicabpviz5409n6dbgtffdmohhn.png)
Therefore, the solutions are
![x_(1) =-6\\x_(2) =1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/t489zqc6xp4fnn71gtk2wo6izp68fbh9i0.png)