If the value of $b=\sqrt{7}$, evaluate the expression ${(b+2)}^2+{(b-2)}^2$.

The correct answer is 22.

We are given that:
$b=\sqrt{7}$
We need to find the value of the algebraic expression:
${(b+2)}^2+{(b-2)}^2$

Step 1: Expand and simplify the expression algebraically
Recall the fundamental algebraic identity for the sum of two squared binomials:
${(x+y)}^2+{(x-y)}^2=2(x^2+y^2)$

Expanding each term individually:
${(b+2)}^2=b^2+4b+4$
${(b-2)}^2=b^2-4b+4$

Adding the two expressions together:
${(b+2)}^2+{(b-2)}^2=(b^2+4b+4)+(b^2-4b+4)$

Notice that the positive and negative middle linear terms
$+4b$
and
$-4b$
cancel each other out:
${(b+2)}^2+{(b-2)}^2=2b^2+8$

Step 2: Substitute the given value of b
Since we are given
$b=\sqrt{7}$
squaring both sides yields:
$b^2={\left(\sqrt{7}\right)}^2=7$

Now, substitute
$b^2=7$
into our simplified expression:
$2b^2+8=2\left(7\right)+8$
$=14+8$
$=22$

Therefore, the evaluated result of the expression is 22.