Solve 3[(5-2)^2-(4-1)^2+2]+(6+4^2)

$$3 [ ( 5 - 2 ) ^ { 2 } - ( 4 - 1 ) ^ { 2 } + 2 ] + ( 6 + 4 ^ { 2 } )$$

Evaluate

$$28$$

Solution Steps

Subtract 2 from 5 to get 3.

$$3\left(3^{2}-\left(4-1\right)^{2}+2\right)+6+4^{2}$$

Calculate 3 to the power of 2 and get 9.

$$3\left(9-\left(4-1\right)^{2}+2\right)+6+4^{2}$$

Subtract 1 from 4 to get 3.

$$3\left(9-3^{2}+2\right)+6+4^{2}$$

Calculate 3 to the power of 2 and get 9.

$$3\left(9-9+2\right)+6+4^{2}$$

Subtract 9 from 9 to get 0.

$$3\left(0+2\right)+6+4^{2}$$

Add 0 and 2 to get 2.

$$3\times 2+6+4^{2}$$

Multiply 3 and 2 to get 6.

$$6+6+4^{2}$$

Add 6 and 6 to get 12.

$$12+4^{2}$$

Calculate 4 to the power of 2 and get 16.

$$12+16$$

Add 12 and 16 to get 28.

$$28$$

Factor

$$2^{2}\times 7$$