What are the solutions to the equation -x^2+2x+48=0 ?

The solutions to the equation -x^2+2x+48=0 are x=-6,x=8

Find the zeros of -x^2+2x+48=0

The zeros of -x^2+2x+48=0 are x=-6,x=8

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- x^{2} + 2 x + 48 = 0

x = - 6, x = 8

Solution steps

- x^{2} + 2 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) \cdot 48}{2 ( - 1 )}

2^{2} - 4 (- 1) \cdot 48 = 14

x_{1, 2} = \frac{- 2 \pm 14}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 2 + 14}{2 ( - 1 )}, x_{2} = \frac{- 2 - 14}{2 ( - 1 )}

x = \frac{- 2 + 14}{2 ( - 1 )} : - 6

x = \frac{- 2 - 14}{2 ( - 1 )} : 8

The solutions to the quadratic equation are:

x = - 6, x = 8

x (x - 3) - 2 (x - 3) = 6

2 (w - 2) (w + 1) - (w + 3) = 0

5000 x^{2} + 5000 x - 9254.27 = 0

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

1.5 t - 0.12 t^{2} = 0

What are the solutions to the equation -x^2+2x+48=0 ?
The solutions to the equation -x^2+2x+48=0 are x=-6,x=8
Find the zeros of -x^2+2x+48=0
The zeros of -x^2+2x+48=0 are x=-6,x=8