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

The solutions to the equation 112=-16x^2+128x are x=1,x=7

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

The zeros of 112=-16x^2+128x are x=1,x=7

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

x = 1, x = 7

Solution steps

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

Switch sides

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

Move

112

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- 128 \pm 12 8 ^{2} - 4 ( - 16 ) ( - 112 )}{2 ( - 16 )}

12 8^{2} - 4 (- 16) (- 112) = 96

x_{1, 2} = \frac{- 128 \pm 96}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 128 + 96}{2 ( - 16 )}, x_{2} = \frac{- 128 - 96}{2 ( - 16 )}

x = \frac{- 128 + 96}{2 ( - 16 )} : 1

x = \frac{- 128 - 96}{2 ( - 16 )} : 7

The solutions to the quadratic equation are:

x = 1, x = 7

- 49 p^{2} = 7

- 2 x + x^{2} + \frac{3}{4} = 0

(2 x + 1) (x - 8) = 0

3 x^{2} - 4 x = 20 x + 27

k = r^{2} + 5, 6 \leq r \leq 7

What are the solutions to the equation 112=-16x^2+128x ?
The solutions to the equation 112=-16x^2+128x are x=1,x=7
Find the zeros of 112=-16x^2+128x
The zeros of 112=-16x^2+128x are x=1,x=7