Solve frac{(3+4)^2-sqrt{64}}{1^6-4^2}

$$\frac { ( 3 + 4 ) ^ { 2 } - \sqrt { 64 } } { 1 ^ { 6 } - 4 ^ { 2 } }$$

Evaluate

$$-\frac{41}{15}\approx -2.733333333$$

Solution Steps

Add 3 and 4 to get 7.

$$\frac{7^{2}-\sqrt{64}}{1^{6}-4^{2}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{49-\sqrt{64}}{1^{6}-4^{2}}$$

Calculate the square root of 64 and get 8.

$$\frac{49-8}{1^{6}-4^{2}}$$

Subtract 8 from 49 to get 41.

$$\frac{41}{1^{6}-4^{2}}$$

Calculate 1 to the power of 6 and get 1.

$$\frac{41}{1-4^{2}}$$

Calculate 4 to the power of 2 and get 16.

$$\frac{41}{1-16}$$

Subtract 16 from 1 to get -15.

$$\frac{41}{-15}$$

Fraction \frac{41}{-15} can be rewritten as -\frac{41}{15} by extracting the negative sign.

$$-\frac{41}{15}$$

Factor

$$-\frac{41}{15} = -2\frac{11}{15} = -2.7333333333333334$$