Evaluate square root of 64+(3^2+(9-4(6-1))+5)

$\sqrt{64} + (3^{2} + (9 - 4 (6 - 1)) + 5)$

Remove parentheses.

$\sqrt{64} + 3^{2} + 9 - 4 (6 - 1) + 5$

Simplify each term.

Rewrite $64$ as $8^{2}$ .

$\sqrt{8^{2}} + 3^{2} + 9 - 4 (6 - 1) + 5$

Pull terms out from under the radical, assuming positive real numbers.

$8 + 3^{2} + 9 - 4 (6 - 1) + 5$

Raise $3$ to the power of $2$ .

$8 + 9 + 9 - 4 (6 - 1) + 5$

Subtract $1$ from $6$ .

$8 + 9 + 9 - 4 \cdot 5 + 5$

Multiply $- 4$ by $5$ .

$8 + 9 + 9 - 20 + 5$

Simplify by adding and subtracting.

Add $8$ and $9$ .

$17 + 9 - 20 + 5$

Add $17$ and $9$ .

$26 - 20 + 5$

Subtract $20$ from $26$ .

$6 + 5$

Add $6$ and $5$ .

$11$