What are the solutions to the equation -16x^2+112x+128=0 ?

The solutions to the equation -16x^2+112x+128=0 are x=-1,x=8

Find the zeros of -16x^2+112x+128=0

The zeros of -16x^2+112x+128=0 are x=-1,x=8

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- 16 x^{2} + 112 x + 128 = 0

x = - 1, x = 8

Solution steps

- 16 x^{2} + 112 x + 128 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 112 \pm 11 2 ^{2} - 4 ( - 16 ) \cdot 128}{2 ( - 16 )}

11 2^{2} - 4 (- 16) \cdot 128 = 144

x_{1, 2} = \frac{- 112 \pm 144}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 112 + 144}{2 ( - 16 )}, x_{2} = \frac{- 112 - 144}{2 ( - 16 )}

x = \frac{- 112 + 144}{2 ( - 16 )} : - 1

x = \frac{- 112 - 144}{2 ( - 16 )} : 8

The solutions to the quadratic equation are:

x = - 1, x = 8

n = t^{2} - 12

x^{2} - 2 = 7 x - x^{2}

x^{2} + 12 x - 72 = 0

3 x^{2} + (k - 6) x + 24 = 0

1 5^{2} + 1 5^{2} = c^{2}

What are the solutions to the equation -16x^2+112x+128=0 ?
The solutions to the equation -16x^2+112x+128=0 are x=-1,x=8
Find the zeros of -16x^2+112x+128=0
The zeros of -16x^2+112x+128=0 are x=-1,x=8