What are the solutions to the equation 3x^2+12x-39=0 ?

The solutions to the equation 3x^2+12x-39=0 are x=-2+sqrt(17),x=-2-sqrt(17)

Find the zeros of 3x^2+12x-39=0

The zeros of 3x^2+12x-39=0 are x=-2+sqrt(17),x=-2-sqrt(17)

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3 x^{2} + 12 x - 39 = 0

x = - 2 + 17, x = - 2 - 17

Solution steps

3 x^{2} + 12 x - 39 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 ( - 39 )}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 (- 39) = 617

x_{1, 2} = \frac{- 12 \pm 6 17}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 6 17}{2 \cdot 3}, x_{2} = \frac{- 12 - 6 17}{2 \cdot 3}

x = \frac{- 12 + 6 17}{2 \cdot 3} : - 2 + 17

x = \frac{- 12 - 6 17}{2 \cdot 3} : - 2 - 17

The solutions to the quadratic equation are:

x = - 2 + 17, x = - 2 - 17

\frac{3 ( 2 t + 2 )}{2} > \frac{t - 3}{4} + \frac{t}{3}

6 < - 2 y - 8

2 x - (6 x + 7) > - 19

p^{2} - 19 p + 84 = 0

6 x + 2 + 6 x < 14

What are the solutions to the equation 3x^2+12x-39=0 ?
The solutions to the equation 3x^2+12x-39=0 are x=-2+sqrt(17),x=-2-sqrt(17)
Find the zeros of 3x^2+12x-39=0
The zeros of 3x^2+12x-39=0 are x=-2+sqrt(17),x=-2-sqrt(17)