Question

4x²+3x-4=0 ajudeeeeeem ai genteeeee

Answer 1

$4x^2 + 3x - 4 = 0 \\ ax^2 + bx + c = 0 \\\\ a = 4 \\ b = 3 \\ c = -4 \\\\ \Delta = b^2-4ac \\ \Delta = 3^2 - 4.4.(-4) \\ \Delta = 9 + 64 \\ \Delta = 73$

$x^+ = \frac{-b+ \sqrt{\Delta}}{2a} \\\\ x^+ = \frac{-3 \pm \sqrt{73} }{2.4} \\\\ x^+ = \frac{-3 + \sqrt{73} }{8} ~ \qquad \qquad\quad ~ x^-= \frac{-3 - \sqrt{73}}{8}$