What are the solutions to the equation -16x^2+16x+32=0 ?

The solutions to the equation -16x^2+16x+32=0 are x=-1,x=2

Find the zeros of -16x^2+16x+32=0

The zeros of -16x^2+16x+32=0 are x=-1,x=2

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- 16 x^{2} + 16 x + 32 = 0

x = - 1, x = 2

Solution steps

- 16 x^{2} + 16 x + 32 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 16 ) \cdot 32}{2 ( - 16 )}

1 6^{2} - 4 (- 16) \cdot 32 = 48

x_{1, 2} = \frac{- 16 \pm 48}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 16 + 48}{2 ( - 16 )}, x_{2} = \frac{- 16 - 48}{2 ( - 16 )}

x = \frac{- 16 + 48}{2 ( - 16 )} : - 1

x = \frac{- 16 - 48}{2 ( - 16 )} : 2

The solutions to the quadratic equation are:

x = - 1, x = 2

(1 + \frac{0.07}{12})^{12}

2 \leq \frac{3}{2} x - 1 < 8

2 (4 - x) = 3 (2 - x)

3 + \frac{2 x}{3} = - 3

- 2 x^{3} + x^{2} + 13 x + 6 < 0

What are the solutions to the equation -16x^2+16x+32=0 ?
The solutions to the equation -16x^2+16x+32=0 are x=-1,x=2
Find the zeros of -16x^2+16x+32=0
The zeros of -16x^2+16x+32=0 are x=-1,x=2