Question

bhaskara de 3x^2 + 2x -1

Answer 1

3x²+2x-1=0
a=3
b=2
c=-1
Δ=b²-4·a·c
Δ=2²-4·3·(-1)
Δ=4+12
Δ=16

$x= \frac{-b+ou- \sqrt{ (delta) } }{2a}$

$x= \frac{-2+ou- \sqrt{16} }{2*3}$

$x'= \frac{-2+4}{6} = \frac{2}{6} = \frac{1}{3}$

$x"= \frac{-2-4}{6} = \frac{-6}{6} = -1$
x=(1/3,-1)