Solve b=sqrt{49}-sqrt{108}+sqrt{12}

$$b = \sqrt { 49 } - \sqrt { 108 } + \sqrt { 12 }$$

Solve for b

$$b=7-4\sqrt{3}\approx 0.07179677$$

Steps for Solving Linear Equation

Calculate the square root of 49 and get 7.

$$b=7-\sqrt{108}+\sqrt{12}$$

Factor 108=6^{2}\times 3. Rewrite the square root of the product \sqrt{6^{2}\times 3} as the product of square roots \sqrt{6^{2}}\sqrt{3}. Take the square root of 6^{2}.

$$b=7-6\sqrt{3}+\sqrt{12}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$b=7-6\sqrt{3}+2\sqrt{3}$$

Combine -6\sqrt{3} and 2\sqrt{3} to get -4\sqrt{3}.

$$b=7-4\sqrt{3}$$

Assign b

$$b≔7-4\sqrt{3}$$