What are the solutions to the equation -4x^2+8x+12=0 ?

The solutions to the equation -4x^2+8x+12=0 are x=-1,x=3

Find the zeros of -4x^2+8x+12=0

The zeros of -4x^2+8x+12=0 are x=-1,x=3

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- 4 x^{2} + 8 x + 12 = 0

x = - 1, x = 3

Solution steps

- 4 x^{2} + 8 x + 12 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 4 ) \cdot 12}{2 ( - 4 )}

8^{2} - 4 (- 4) \cdot 12 = 16

x_{1, 2} = \frac{- 8 \pm 16}{2 ( - 4 )}

Separate the solutions

x_{1} = \frac{- 8 + 16}{2 ( - 4 )}, x_{2} = \frac{- 8 - 16}{2 ( - 4 )}

x = \frac{- 8 + 16}{2 ( - 4 )} : - 1

x = \frac{- 8 - 16}{2 ( - 4 )} : 3

The solutions to the quadratic equation are:

x = - 1, x = 3

4 p^{2} + p - 6 = 0

3 x^{2} + 16 x + 1 = 9 x - 3

- 6 x^{2} = - 24

9 k^{2} - 18 k + 5 = 0

x, m x^{2} - m x + 3 = 0

What are the solutions to the equation -4x^2+8x+12=0 ?
The solutions to the equation -4x^2+8x+12=0 are x=-1,x=3
Find the zeros of -4x^2+8x+12=0
The zeros of -4x^2+8x+12=0 are x=-1,x=3