What are the solutions to the equation 3x^2+12x-288=0 ?

The solutions to the equation 3x^2+12x-288=0 are x=8,x=-12

Find the zeros of 3x^2+12x-288=0

The zeros of 3x^2+12x-288=0 are x=8,x=-12

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3 x^{2} + 12 x - 288 = 0

x = 8, x = - 12

Solution steps

3 x^{2} + 12 x - 288 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 \cdot 3 ( - 288 )}{2 \cdot 3}

1 2^{2} - 4 \cdot 3 (- 288) = 60

x_{1, 2} = \frac{- 12 \pm 60}{2 \cdot 3}

Separate the solutions

x_{1} = \frac{- 12 + 60}{2 \cdot 3}, x_{2} = \frac{- 12 - 60}{2 \cdot 3}

x = \frac{- 12 + 60}{2 \cdot 3} : 8

x = \frac{- 12 - 60}{2 \cdot 3} : - 12

The solutions to the quadratic equation are:

x = 8, x = - 12

\frac{f}{5} + 2 = 8

(- 3 m^{5}) (- 2 m^{4})

6 + \frac{2 x + 4}{3} x = 8

2 z^{2} + 13 z + 11

356

What are the solutions to the equation 3x^2+12x-288=0 ?
The solutions to the equation 3x^2+12x-288=0 are x=8,x=-12
Find the zeros of 3x^2+12x-288=0
The zeros of 3x^2+12x-288=0 are x=8,x=-12