What are the solutions to the equation 12x^2+24x-96=0 ?

The solutions to the equation 12x^2+24x-96=0 are x=2,x=-4

Find the zeros of 12x^2+24x-96=0

The zeros of 12x^2+24x-96=0 are x=2,x=-4

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12 x^{2} + 24 x - 96 = 0

x = 2, x = - 4

Solution steps

12 x^{2} + 24 x - 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 24 \pm 2 4 ^{2} - 4 \cdot 12 ( - 96 )}{2 \cdot 12}

2 4^{2} - 4 \cdot 12 (- 96) = 72

x_{1, 2} = \frac{- 24 \pm 72}{2 \cdot 12}

Separate the solutions

x_{1} = \frac{- 24 + 72}{2 \cdot 12}, x_{2} = \frac{- 24 - 72}{2 \cdot 12}

x = \frac{- 24 + 72}{2 \cdot 12} : 2

x = \frac{- 24 - 72}{2 \cdot 12} : - 4

The solutions to the quadratic equation are:

x = 2, x = - 4

5 x^{2} - 5 x + 7 = 0

14 x^{2} - 28 x = 0

x^{2} - 23 x + 90 = 0

(x - 6)^{2} = - 36

x^{2} = (x - 1)^{2} + (x - 8)^{2}

What are the solutions to the equation 12x^2+24x-96=0 ?
The solutions to the equation 12x^2+24x-96=0 are x=2,x=-4
Find the zeros of 12x^2+24x-96=0
The zeros of 12x^2+24x-96=0 are x=2,x=-4