evangelina69
evangelina69 evangelina69
  • 02-11-2020
  • Mathematics
contestada

How do you solve 6x^2+40x-14=0

Respuesta :

Аноним Аноним
  • 02-11-2020

Answer:

x=1/3 (in fraction form)

or x=-7

Step-by-step explanation:

Answer Link
cliffordisodd
cliffordisodd cliffordisodd
  • 02-11-2020

Answer:

x1=0.333, x2=-7

Step-by-step explanation:

[tex]6 {x}^{2} + 40x - 14 = 0 \\ \\ divide \: through \: by \: 2 \\ 3 {x}^{2} + 20 x - 7 = 0 \\ write \: 20x \: as \: a \: difference \\ 3x + 21x - x - 7 = 0 \\ (3 {x}^{2} + 21x) - (x + 7) = 0\\ 3x(x + 7) - (x + 7) = 0 \\ (3x - 1)(x + 7) = 0 \\ 3x - 1 = 0 \\ 3x = 1 \\ x = \frac{1}{3} \\ \\ x + 7 = 0 \\ x = - 7[/tex]

Answer Link

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