What are the roots of the equation \(x^2+14x+10=0\)?

Correct answer: \(x=-7\pm\sqrt{39}\)

Rewrite the quadratic by completing the square: \(x^2+14x+10=(x+7)^2-39\). Thus, \((x+7)^2=39\), so \(x+7=\pm\sqrt{39}\) and \(x=-7\pm\sqrt{39}\). Option B has the wrong sign for the constant term in the root expression, while option D omits the square root. In an exam, remember that \(\pm\) represents both roots.