Solve sqrt{64}-(27:4+24:6)+2^3-3^2=

$$\sqrt { 64 } - ( 27 : 4 + 24 : 6 ) + 2 ^ { 3 } - 3 ^ { 2 } =$$

Evaluate

$$-\frac{15}{4}=-3.75$$

Solution Steps

Calculate the square root of 64 and get 8.

$$8-\left(\frac{27}{4}+\frac{24}{6}\right)+2^{3}-3^{2}$$

Divide 24 by 6 to get 4.

$$8-\left(\frac{27}{4}+4\right)+2^{3}-3^{2}$$

Convert 4 to fraction \frac{16}{4}.

$$8-\left(\frac{27}{4}+\frac{16}{4}\right)+2^{3}-3^{2}$$

Since \frac{27}{4} and \frac{16}{4} have the same denominator, add them by adding their numerators.

$$8-\frac{27+16}{4}+2^{3}-3^{2}$$

Add 27 and 16 to get 43.

$$8-\frac{43}{4}+2^{3}-3^{2}$$

Convert 8 to fraction \frac{32}{4}.

$$\frac{32}{4}-\frac{43}{4}+2^{3}-3^{2}$$

Since \frac{32}{4} and \frac{43}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{32-43}{4}+2^{3}-3^{2}$$

Subtract 43 from 32 to get -11.

$$-\frac{11}{4}+2^{3}-3^{2}$$

Calculate 2 to the power of 3 and get 8.

$$-\frac{11}{4}+8-3^{2}$$

Convert 8 to fraction \frac{32}{4}.

$$-\frac{11}{4}+\frac{32}{4}-3^{2}$$

Since -\frac{11}{4} and \frac{32}{4} have the same denominator, add them by adding their numerators.

$$\frac{-11+32}{4}-3^{2}$$

Add -11 and 32 to get 21.

$$\frac{21}{4}-3^{2}$$

Calculate 3 to the power of 2 and get 9.

$$\frac{21}{4}-9$$

Convert 9 to fraction \frac{36}{4}.

$$\frac{21}{4}-\frac{36}{4}$$

Since \frac{21}{4} and \frac{36}{4} have the same denominator, subtract them by subtracting their numerators.

$$\frac{21-36}{4}$$

Subtract 36 from 21 to get -15.

$$-\frac{15}{4}$$

Factor

$$-\frac{15}{4} = -3\frac{3}{4} = -3.75$$