Solve 3-sqrt{49}+2(3+4^2)div12

$$3 - \sqrt { 49 } + 2 ( 3 + 4 ^ { 2 } ) \div 12$$

Evaluate

$$-\frac{5}{6}\approx -0.833333333$$

Solution Steps

Calculate the square root of 49 and get 7.

$$3-7+\frac{2\left(3+4^{2}\right)}{12}$$

Subtract 7 from 3 to get -4.

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

Calculate 4 to the power of 2 and get 16.

$$-4+\frac{2\left(3+16\right)}{12}$$

Add 3 and 16 to get 19.

$$-4+\frac{2\times 19}{12}$$

Multiply 2 and 19 to get 38.

$$-4+\frac{38}{12}$$

Reduce the fraction \frac{38}{12} to lowest terms by extracting and canceling out 2.

$$-4+\frac{19}{6}$$

Convert -4 to fraction -\frac{24}{6}.

$$-\frac{24}{6}+\frac{19}{6}$$

Since -\frac{24}{6} and \frac{19}{6} have the same denominator, add them by adding their numerators.

$$\frac{-24+19}{6}$$

Add -24 and 19 to get -5.

$$-\frac{5}{6}$$

Factor

$$-\frac{5}{6} = -0.8333333333333334$$