## Algebra: A Combined Approach (4th Edition)

$2\sqrt{7}$
Given $a = 1, b = 6, c = 2$ we have $\sqrt{b^2-4ac} = \sqrt{6^2-4 \times 1 \times 2} = \sqrt{36-8} = \sqrt{28} = \sqrt{4 \times 7} = \sqrt{2^2 \times 7} = 2\sqrt{7}$