Question

resolva com a fórmula de Bhaskara:2x^2+40x=0​

Answer 1

$2 {x}^{2} + 40x = 0$

$a {x}^{2} + bx + c = 0$

$a = 2$

$b = 40$

$c = 0$

$x = \frac{ - b + - \sqrt{ {b}^{2} - 4ac } }{2a}$


$x = \frac{ - 40 + - \sqrt{ {40}^{2} - 4 \times 2 \times 0 } }{2 \times 2}$

$x = \frac{ - 40 + - \sqrt{ {40}^{2} } }{4}$

$x = \frac{ - 40 + - 40}{4}$


${x}^{.} = \frac{ - 40 + 40}{4} = 0$

${x}^{..} = \frac{ - 40 - 40}{4} = \frac{ - 80}{4} = - 20$