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

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

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

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

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

x = - 2, x = 3

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 (- 2) \cdot 12 = 10

x_{1, 2} = \frac{- 2 \pm 10}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 2 + 10}{2 ( - 2 )}, x_{2} = \frac{- 2 - 10}{2 ( - 2 )}

x = \frac{- 2 + 10}{2 ( - 2 )} : - 2

x = \frac{- 2 - 10}{2 ( - 2 )} : 3

The solutions to the quadratic equation are:

x = - 2, x = 3

4 x^{2} + 256 = 0

- 3 x^{2} + 12 x + 429 = 0

4 (x - 2)^{2} = - 1

2 x^{2} = 22

2 x^{2} = 7 x

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