Solve 8^2-2*7^2/-6^2+(-2)^2

$$\frac { 8 ^ { 2 } - 2 \cdot 7 ^ { 2 } } { - 6 ^ { 2 } + ( - 2 ) ^ { 2 } }$$

Evaluate

$$\frac{17}{16}=1.0625$$

Solution Steps

Calculate 8 to the power of 2 and get 64.

$$\frac{64-2\times 7^{2}}{-6^{2}+\left(-2\right)^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{64-2\times 49}{-6^{2}+\left(-2\right)^{2}}$$

Multiply 2 and 49 to get 98.

$$\frac{64-98}{-6^{2}+\left(-2\right)^{2}}$$

Subtract 98 from 64 to get -34.

$$\frac{-34}{-6^{2}+\left(-2\right)^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$\frac{-34}{-36+\left(-2\right)^{2}}$$

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

$$\frac{-34}{-36+4}$$

Add -36 and 4 to get -32.

$$\frac{-34}{-32}$$

Reduce the fraction \frac{-34}{-32} to lowest terms by extracting and canceling out -2.

$$\frac{17}{16}$$

Factor

$$\frac{17}{2 ^ {4}} = 1\frac{1}{16} = 1.0625$$