What are the solutions to the equation 48+8x-x^2=0 ?

The solutions to the equation 48+8x-x^2=0 are x=-4,x=12

Find the zeros of 48+8x-x^2=0

The zeros of 48+8x-x^2=0 are x=-4,x=12

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48 + 8 x - x^{2} = 0

x = - 4, x = 12

Solution steps

48 + 8 x - x^{2} = 0

Write in the standard form

a x^{2} + b x + c = 0

- x^{2} + 8 x + 48 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 1 ) \cdot 48}{2 ( - 1 )}

8^{2} - 4 (- 1) \cdot 48 = 16

x_{1, 2} = \frac{- 8 \pm 16}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 8 + 16}{2 ( - 1 )}, x_{2} = \frac{- 8 - 16}{2 ( - 1 )}

x = \frac{- 8 + 16}{2 ( - 1 )} : - 4

x = \frac{- 8 - 16}{2 ( - 1 )} : 12

The solutions to the quadratic equation are:

x = - 4, x = 12

r^{2} - 5 r = 0

- x^{2} + 5 x - 2 = 0

2 x (x + 1) = 0

(x - 5) (x + 4) = 0

x^{2} + 5 x + 2 = 0

What are the solutions to the equation 48+8x-x^2=0 ?
The solutions to the equation 48+8x-x^2=0 are x=-4,x=12
Find the zeros of 48+8x-x^2=0
The zeros of 48+8x-x^2=0 are x=-4,x=12