\(3x^2 - 12x + 24 = 0 \\ x^2 - 4x + 8 = 0 \\ x= \frac{-b \pm \sqrt{ b^{2} -4ac} }{2a}\); where a = 1, b = -4 and c = 8
\(x= \frac{-(-4) \pm \sqrt{ (-4)^{2} -4 \times 1 \times 8} }{2 \times 1} \\ = \frac{4 \pm \sqrt{ 16 -32} }{2} \\ =\frac{4 \pm \sqrt{ -16} }{2} \\ =\frac{4 \pm 4i }{2} \\ =2+2i \ or \ 2-2i\)
