Solve (3^{2}+5*2^2)+(5*3-3)^2*3^2-7^0=

$$( 3 ^ { 2 } + 5 \times 2 ^ { 2 } ) + ( 5 \times 3 - 3 ) ^ { 2 } \times 3 ^ { 2 } - 7 ^ { 0 } =$$

Evaluate

$$1324$$

Solution Steps

Calculate 3 to the power of 2 and get 9.

$$9+5\times 2^{2}+\left(5\times 3-3\right)^{2}\times 3^{2}-7^{0}$$

Calculate 2 to the power of 2 and get 4.

$$9+5\times 4+\left(5\times 3-3\right)^{2}\times 3^{2}-7^{0}$$

Multiply 5 and 4 to get 20.

$$9+20+\left(5\times 3-3\right)^{2}\times 3^{2}-7^{0}$$

Add 9 and 20 to get 29.

$$29+\left(5\times 3-3\right)^{2}\times 3^{2}-7^{0}$$

Multiply 5 and 3 to get 15.

$$29+\left(15-3\right)^{2}\times 3^{2}-7^{0}$$

Subtract 3 from 15 to get 12.

$$29+12^{2}\times 3^{2}-7^{0}$$

Calculate 12 to the power of 2 and get 144.

$$29+144\times 3^{2}-7^{0}$$

Calculate 3 to the power of 2 and get 9.

$$29+144\times 9-7^{0}$$

Multiply 144 and 9 to get 1296.

$$29+1296-7^{0}$$

Add 29 and 1296 to get 1325.

$$1325-7^{0}$$

Calculate 7 to the power of 0 and get 1.

$$1325-1$$

Subtract 1 from 1325 to get 1324.

$$1324$$

Factor

$$2^{2}\times 331$$