Question

podem me ajudar nesta fórmula de baskhara?!​

Answer 1

Resposta:

$x' = \frac{1}{5}$

$x'' = -1$

Explicação passo-a-passo:

$20.x^{2} + 11x - 3 = 0$

Usando Bhaskara

$x' = \frac{-b + \sqrt{b^{2} - 4.a.c } }{2.a}$

$x' = \frac{-11 + \sqrt{11^{2} - (4.20.-3)} }{2.20}$

$x' = \frac{-11 + \sqrt{121 - (-240)} }{40}$

$x' = \frac{-11 + \sqrt{121 + 240} }{40}$

$x' = \frac{-11 + \sqrt{361} }{40}$

$x' = \frac{-11 + \sqrt{19^2} }{40}$

$x' = \frac{-11 + 19 }{40}$

$x' = \frac{8}{40} = \frac{1}{5}$

$x'' = \frac{-11 - 19 }{40}$

$x'' = \frac{-40}{40} = -1$