Solve 12+sqrt{2^2}-frac{sqrt{6}}{3}

$$12 + \sqrt { 2 ^ { 2 } } - \frac { \sqrt { 6 } } { 3 }$$

Evaluate

$$-\frac{\sqrt{6}}{3}+14\approx 13.183503419$$

Short Solution Steps

Calculate 2 to the power of 2 and get 4.

$$12+\sqrt{4}-\frac{\sqrt{6}}{3}$$

Calculate the square root of 4 and get 2.

$$12+2-\frac{\sqrt{6}}{3}$$

Add 12 and 2 to get 14.

$$14-\frac{\sqrt{6}}{3}$$

To add or subtract expressions, expand them to make their denominators the same. Multiply 14 times \frac{3}{3}.

$$\frac{14\times 3}{3}-\frac{\sqrt{6}}{3}$$

Since \frac{14\times 3}{3} and \frac{\sqrt{6}}{3} have the same denominator, subtract them by subtracting their numerators.

$$\frac{14\times 3-\sqrt{6}}{3}$$

Do the multiplications in 14\times 3-\sqrt{6}.

$$\frac{42-\sqrt{6}}{3}$$

Factor

$$\frac{42 - \sqrt{6}}{3} = 13.183503419072274$$