Question

resolva a equaçao 2x²+8x+6=0

Answer 1

2x²+8x+6=0, utilizamos Bhaskara

x= $\frac{-b+- \sqrt{b^2-4ac} }{2a}$

onde Δ=b²-4ac⇒Δ=8²-4.2.6 = 64-48 = 16

x= -8₋⁺√16 / 2.2

x1 = -8 + 4 / 4 = -4/4 = -1

x2 = -8 -4 / 4 = -12/4 = -3

Answer 2

Vamos resolver pela Fórmula de Bhaskara, então, será necessário determinar os coeficientes:
$2x^{2}+8x+6=0$

Coeficientes:
a=2
b=8
c=6

Calculando Δ:
Δ=b²-4ac
Δ=8²-4*2*6
Δ=64-48
Δ=16

$x= \dfrac{-b+- \sqrt{16}}{2a}= \dfrac{-8+-4}{2*2}=\dfrac{-8+-4}{4}$

$x'=\dfrac{-8+4}{4}= \dfrac{-4}{4}=-1$

$x''=\dfrac{-8-4}{4}=\dfrac{-12}{4}=-3$

$S=(-1,-3)$

Espero ter ajudado :D