Question

Formula de Bháskara!

$x^{2} - \frac{19}{15} x + \frac{2}{5} = 0$

Por favor, me ajudem!

Answer 1

$x^{2} -\frac{19}{15}x+\frac{2}{5}=0\\ \\ 15 x^{2} -19x+10=0\\ \\ \Delta=(-19)^2-4.15.2=361-120=241\\ \\ \boxed{x=\frac{19 \pm\sqrt{241}}{30}}$

Answer 2

      mmc = 15

15x² - 19x + 6 = 0

Δ= (-19)² - 4.15.6==> 361-360==> 1

x = 19+/-√1 ==> x = 19+/-1
          2.1                    2

x1 = 19+1 ==> x1 = 10
          2

x2 = 19-1 ==> x2 = 9
          2