Question

10x²+3x-1=0
quais são as raízes dessa equação ?

Answer 1

$10x^2 + 3x - 1 = 0$

Δ=b²−4ac
Δ=(3)2−4⋅(10)⋅(−1)
Δ=9+4
Δ=49

===

$x = \dfrac{-b \pm \sqrt{\triangle}}{2a} \\ \\ x = \dfrac{-3 \pm \sqrt{49}}{2 * 10} \\ \\ x = \dfrac{-3 \pm 7}{20} \\ \\ x' = \dfrac{-3 +7}{6} \\ \\ x' = \dfrac{4}{20} \\ \\ x' = \dfrac{1}{5}$


$x'' = \dfrac{-3 - 7}{6} \\ \\ x'' = \dfrac{-10}{20} \\ \\ x''= -\dfrac{1}{2}$


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

Answer 2

10x²+3x-1=0
∆=b²-4ac
∆=3²-4.10.(-1)
∆=9+40
∆=49
x=-b±√∆/2a
x=-3-7/2.10
x=-10/20
x=-1/2
x=-3+7/2.10
x=4/20 (÷4)
x=1/5