Solve -36:3^2+sqrt[2]{289}-sqrt[2]{sqrt[3]{125}+(-2)^2}+7*2=

$$- 36 : 3 ^ { 2 } + \sqrt[ 2 ] { 289 } - \sqrt[ 2 ] { \sqrt[ 3 ] { 125 } + ( - 2 ) ^ { 2 } } + 7 \cdot 2 =$$

Evaluate

$$24$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$\frac{-36}{9}+\sqrt[2]{289}-\sqrt[2]{\sqrt[3]{125}+\left(-2\right)^{2}}+7\times 2$$

Divide -36 by 9 to get -4.

$$-4+\sqrt[2]{289}-\sqrt[2]{\sqrt[3]{125}+\left(-2\right)^{2}}+7\times 2$$

Calculate \sqrt[2]{289} and get 17.

$$-4+17-\sqrt[2]{\sqrt[3]{125}+\left(-2\right)^{2}}+7\times 2$$

Add -4 and 17 to get 13.

$$13-\sqrt[2]{\sqrt[3]{125}+\left(-2\right)^{2}}+7\times 2$$

Calculate \sqrt[3]{125} and get 5.

$$13-\sqrt[2]{5+\left(-2\right)^{2}}+7\times 2$$

Calculate -2 to the power of 2 and get 4.

$$13-\sqrt[2]{5+4}+7\times 2$$

Add 5 and 4 to get 9.

$$13-\sqrt[2]{9}+7\times 2$$

Calculate \sqrt[2]{9} and get 3.

$$13-3+7\times 2$$

Subtract 3 from 13 to get 10.

$$10+7\times 2$$

Multiply 7 and 2 to get 14.

$$10+14$$

Add 10 and 14 to get 24.

$$24$$

Factor

$$2^{3}\times 3$$