Solve 8-3*(4-2)^2+12:(3*2-4)^3

$$8 - 3 \cdot ( 4 - 2 ) ^ { 2 } + 12 : ( 3 \cdot 2 - 4 ) ^ { 3 }$$

Evaluate

$$-\frac{5}{2}=-2.5$$

Solution Steps

Subtract 2 from 4 to get 2.

$$8-3\times 2^{2}+\frac{12}{\left(3\times 2-4\right)^{3}}$$

Calculate 2 to the power of 2 and get 4.

$$8-3\times 4+\frac{12}{\left(3\times 2-4\right)^{3}}$$

Multiply 3 and 4 to get 12.

$$8-12+\frac{12}{\left(3\times 2-4\right)^{3}}$$

Subtract 12 from 8 to get -4.

$$-4+\frac{12}{\left(3\times 2-4\right)^{3}}$$

Multiply 3 and 2 to get 6.

$$-4+\frac{12}{\left(6-4\right)^{3}}$$

Subtract 4 from 6 to get 2.

$$-4+\frac{12}{2^{3}}$$

Calculate 2 to the power of 3 and get 8.

$$-4+\frac{12}{8}$$

Reduce the fraction \frac{12}{8} to lowest terms by extracting and canceling out 4.

$$-4+\frac{3}{2}$$

Add -4 and \frac{3}{2} to get -\frac{5}{2}.

$$-\frac{5}{2}$$

Factor

$$-\frac{5}{2} = -2\frac{1}{2} = -2.5$$