Question

Me ajudem a resolver, fórmula de Bhaskara

6x^2 - 2x = 1 -3x

Answer 1

$6x^{2} -2x +3x -1 = 0$

$6x^{2} +1x -1 =0$

Δ=$1^{2} -4.6.-1$

Δ=$\sqrt{25} =$ ±5

$x1 = \frac{-1+5}{2.6} =\frac{4}{12} = \frac{1}{3}$

$x2 = \frac{-1-5}{2.6} = -\frac{6}{12} =-\frac{1}{2}$

Answer 2

Oi

6x² - 2x = 1 - 3x

6x² - 2x + 3x - 1 =0

6x² + x - 1 = 0

a= 6

b= 1

c= - 1

Calculando Delta

Δ = b² - 4.a.c

Δ = 1² - 4.6.-1

Δ = 1 + 24

Δ = 25

Aplicando Bhaskara

x = - b +-√Δ/2.a

x' = - 1 + √25/2.6

x' = - 1 + 5/12

x' = 4/12 ÷ 4

x' = 1/3

x" = -1 -√25/2.6

x" = - 1 - 5/12

x" = - 6 /12 ÷ 6

x" = - 1/2

S={x'=1/3 e x" = - 1/2}

Bons estudos