Question

sa se rezolve in C ecuatia z^2+3z+4=0

Answer 1

   
[tex]\displaystyle\\ z^2+3z+4=0\\\\ z_{12}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}= \frac{-3\pm \sqrt{3^2-4\times 1 \times 4}}{2\times 1}=\\\\ = \frac{-3\pm \sqrt{9-16}}{2}=\frac{-3\pm \sqrt{-7}}{2}=\frac{-3\pm i\sqrt{7}}{2}\\\\ z_1=\frac{-3+ i\sqrt{7}}{2}=\boxed{- \frac{3}{2} + \frac{\sqrt{7}}{2}i} \\\\ z_2=\frac{-3- i\sqrt{7}}{2}=\boxed{- \frac{3}{2} - \frac{\sqrt{7}}{2}i} [/tex]