Answer:
Step-by-step explanation:
Hello, please consider the following.
[tex]-4(b+2)^2+36=0\\\\<=> 36-4(b+2)^2=0\\\\<=> \dfrac{36}{4}-(b+2)^2=0\\\\<=> 3^2-(b+2)^2=(3-b-2)(3+b+2)=\boxed{\sf \bf (1-b)(5+b)=0}\\\\<=> b = 1 \ \ or \ \ b=-5[/tex]
Thank you
