What are the solutions to the equation -2x^2+28x-96=0 ?

The solutions to the equation -2x^2+28x-96=0 are x=6,x=8

Find the zeros of -2x^2+28x-96=0

The zeros of -2x^2+28x-96=0 are x=6,x=8

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- 2 x^{2} + 28 x - 96 = 0

x = 6, x = 8

Solution steps

- 2 x^{2} + 28 x - 96 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 28 \pm 2 8 ^{2} - 4 ( - 2 ) ( - 96 )}{2 ( - 2 )}

2 8^{2} - 4 (- 2) (- 96) = 4

x_{1, 2} = \frac{- 28 \pm 4}{2 ( - 2 )}

Separate the solutions

x_{1} = \frac{- 28 + 4}{2 ( - 2 )}, x_{2} = \frac{- 28 - 4}{2 ( - 2 )}

x = \frac{- 28 + 4}{2 ( - 2 )} : 6

x = \frac{- 28 - 4}{2 ( - 2 )} : 8

The solutions to the quadratic equation are:

x = 6, x = 8

12 x^{2} - 4 x y + 9 x y - 3 y^{2}

x^{3} - y^{3}

4 x = 4 x + 8

2 x + 1 < 3

28 s^{2} + 8 s t - 20 t^{2}

What are the solutions to the equation -2x^2+28x-96=0 ?
The solutions to the equation -2x^2+28x-96=0 are x=6,x=8
Find the zeros of -2x^2+28x-96=0
The zeros of -2x^2+28x-96=0 are x=6,x=8