What are the solutions to the equation 0=-16x^2+32x+128 ?

The solutions to the equation 0=-16x^2+32x+128 are x=-2,x=4

Find the zeros of 0=-16x^2+32x+128

The zeros of 0=-16x^2+32x+128 are x=-2,x=4

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0 = - 16 x^{2} + 32 x + 128

x = - 2, x = 4

Solution steps

0 = - 16 x^{2} + 32 x + 128

Switch sides

- 16 x^{2} + 32 x + 128 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 32 \pm 3 2 ^{2} - 4 ( - 16 ) \cdot 128}{2 ( - 16 )}

3 2^{2} - 4 (- 16) \cdot 128 = 96

x_{1, 2} = \frac{- 32 \pm 96}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 32 + 96}{2 ( - 16 )}, x_{2} = \frac{- 32 - 96}{2 ( - 16 )}

x = \frac{- 32 + 96}{2 ( - 16 )} : - 2

x = \frac{- 32 - 96}{2 ( - 16 )} : 4

The solutions to the quadratic equation are:

x = - 2, x = 4

2 (x + 7)^{2} - 3 x (x + 7)

3 - 2 (n - 4) > - 1

\frac{18 a}{7} \cdot \frac{14 a ^{6}}{3}

(\frac{3}{4})^{10}

- 2 x^{2} + 2 x = 2.5

What are the solutions to the equation 0=-16x^2+32x+128 ?
The solutions to the equation 0=-16x^2+32x+128 are x=-2,x=4
Find the zeros of 0=-16x^2+32x+128
The zeros of 0=-16x^2+32x+128 are x=-2,x=4