What are the solutions to the equation -16x^2+16x+480=0 ?

The solutions to the equation -16x^2+16x+480=0 are x=-5,x=6

Find the zeros of -16x^2+16x+480=0

The zeros of -16x^2+16x+480=0 are x=-5,x=6

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- 16 x^{2} + 16 x + 480 = 0

x = - 5, x = 6

Solution steps

- 16 x^{2} + 16 x + 480 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 16 \pm 1 6 ^{2} - 4 ( - 16 ) \cdot 480}{2 ( - 16 )}

1 6^{2} - 4 (- 16) \cdot 480 = 176

x_{1, 2} = \frac{- 16 \pm 176}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 16 + 176}{2 ( - 16 )}, x_{2} = \frac{- 16 - 176}{2 ( - 16 )}

x = \frac{- 16 + 176}{2 ( - 16 )} : - 5

x = \frac{- 16 - 176}{2 ( - 16 )} : 6

The solutions to the quadratic equation are:

x = - 5, x = 6

- \frac{2 x + 3}{( x - 1 ) ^{2} ( x + 4 ) ^{2}} < 0

0 = (x + 4)^{2} - 2

19 \leq 5 x - 1 \leq 39

∣ x + 3 ∣ > 8

7 x^{2} - 20 x + 4 = 8 x - 10

What are the solutions to the equation -16x^2+16x+480=0 ?
The solutions to the equation -16x^2+16x+480=0 are x=-5,x=6
Find the zeros of -16x^2+16x+480=0
The zeros of -16x^2+16x+480=0 are x=-5,x=6