Given quadratic equation is x2 + 6x + 6 = 0
By comparing given quadratic equation with ax2 + bx + c = 0,
we get a =1, b = 6 and c = 6.
The discriminant of given quadratic equation is
D = b2 − 4ac = 62 − 4 × 1 × 6 = 36 – 24 = 12 > 0 (real and distinct roots)
Hence, the roots of given quadratic equation are real and distinct.
And the roots are \(\frac{-b±\sqrt{D}}{2a}\) = \(\frac{-6±\sqrt{12}}{2\times1}\) = \(\frac{-6±2\sqrt{3}}{2}\) = -3 ± \(\sqrt3\).
Hence, the roots (or solution) of given quadratic equation are x = -3 +√3 and x = -3 − √3.