Question

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

Answer 1

x1=0.333, x2=-7

Explanation:

`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 = (1)/(3) \\ \\ x + 7 = 0 \\ x = - 7`

Answer 2

x=1/3 (in fraction form)

or x=-7

Explanation: