the other one can't be solveed using conventional means use quadratic formula for ax^2+bx+c=0 x=[tex] \frac{-b+/- \sqrt{b^2-4ac} }{2a} [/tex] for x²-4x+16=0 x=[tex] \frac{-(-4)+/- \sqrt{(-4)^2-4(1)(16)} }{2(1)} [/tex] x=[tex] \frac{4+/- \sqrt{16-64} }{2} [/tex] x=[tex] \frac{4+/- \sqrt{-48} }{2} [/tex] x=[tex] \frac{4+/- (\sqrt{-1})(\sqrt{48}) }{2} [/tex] x=[tex] \frac{4+/- (i)(4\sqrt{3}) }{2} [/tex] x=[tex] 2+/- 2i\sqrt{3} [/tex]
