Question

sabendo que a=5,b= -9,c= -2 qual o valor numérico de

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

Answer 1

$x = \frac{-b \pm \sqrt{b^2 -4*a*c}}{2*a} \\ \\ 5x^2 - 9x - 2 = 0$

a=5,b=−9,c=−2

Δ=b2−4ac
Δ=(−9)2−4*(5)*(−2)
Δ=81+40
Δ=121

$x = \frac{-b \pm \sqrt{\triangle}}{2*a} \\ \\ x = \frac{-(-9) \pm \sqrt{121}}{2*5} \\ \\ x = \frac{9 \pm 11}{10} \\ \\ x' = \frac{9 + 11}{10} \\ \\ x' = \frac{20}{10} \\ \\ x' = 2 \\ \\ \\ x'' = \frac{9 - 11}{10}\\ \\x'' = \frac{-2}{10} \\ \\ x'' = \frac{-1}{5}$

S = {$2, \frac{-1}{5}$}