Respuesta :

Lilith
[tex]7x^2+3x+8=0 \\ \\a =7, \ b=3 ,\ c=8 \\ \\\Delta =b^2 - 4ac = 3^2 -4\cdot 7\cdot 8=9-224=-215\\ \\ Answer : \ \Delta \ is \ negative,\ this \ gives \ two \ imaginary \ solutions[/tex]


kate200468
[tex]7x^2+3x+8=0\\\\ \Delta=3^2-4\cdot7\cdot8=9-224=-115\ \ \Rightarrow\ \ \ no\ real\ solution\\\\\ two\ complex\ solutions\\\\\Delta=-115=115\cdot i^2\ \ \ \Rightarrow\ \ \ \sqrt{\Delta} = \sqrt{115}\ i\\\\x_1= \frac{-b- \sqrt{\Delta} }{2a} = \frac{-3- \sqrt{115}\ i}{2\cdot 7} = \frac{-3- \sqrt{115i}}{14}\\\\x_2= \frac{-b+ \sqrt{\Delta} }{2a} = \frac{-3+ \sqrt{115}\ i}{2\cdot 7} = \frac{-3+ \sqrt{115}\ i}{14}[/tex]