Solve A=2^4*3+4^2*3^0+-2^2

$$A = { 2 }^{ 4 } \times 3+ { 4 }^{ 2 } \times { 3 }^{ 0 } + { -2 }^{ 2 }$$

Solve for A

$$A=68$$

Steps for Solving Linear Equation

Calculate 2 to the power of 4 and get 16.

$$A=16\times 3+4^{2}\times 3^{0}+\left(-2\right)^{2}$$

Multiply 16 and 3 to get 48.

$$A=48+4^{2}\times 3^{0}+\left(-2\right)^{2}$$

Calculate 4 to the power of 2 and get 16.

$$A=48+16\times 3^{0}+\left(-2\right)^{2}$$

Calculate 3 to the power of 0 and get 1.

$$A=48+16\times 1+\left(-2\right)^{2}$$

Multiply 16 and 1 to get 16.

$$A=48+16+\left(-2\right)^{2}$$

Add 48 and 16 to get 64.

$$A=64+\left(-2\right)^{2}$$

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

$$A=64+4$$

Add 64 and 4 to get 68.

$$A=68$$

Assign A

$$A≔68$$