What are the solutions to the equation 16-x^2=4+x ?

The solutions to the equation 16-x^2=4+x are x=-4,x=3

Find the zeros of 16-x^2=4+x

The zeros of 16-x^2=4+x are x=-4,x=3

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16 - x^{2} = 4 + x

x = - 4, x = 3

Solution steps

16 - x^{2} = 4 + x

Move

x

to the left side

16 - x^{2} - x = 4

Move

4

to the left side

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

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 12}{2 ( - 1 )}

(- 1)^{2} - 4 (- 1) \cdot 12 = 7

x_{1, 2} = \frac{- ( - 1 ) \pm 7}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 7}{2 ( - 1 )}, x_{2} = \frac{- ( - 1 ) - 7}{2 ( - 1 )}

x = \frac{- ( - 1 ) + 7}{2 ( - 1 )} : - 4

x = \frac{- ( - 1 ) - 7}{2 ( - 1 )} : 3

The solutions to the quadratic equation are:

x = - 4, x = 3

12 h^{2} - 152 h + 325 = 0

- 16 t^{2} + 560 t = 0

16 x^{2} - 120 x + 180 = 0

x^{2} - 15 x + 26 = - 3 x - 6

8 x^{2} + 24 x - 20 = 0

What are the solutions to the equation 16-x^2=4+x ?
The solutions to the equation 16-x^2=4+x are x=-4,x=3
Find the zeros of 16-x^2=4+x
The zeros of 16-x^2=4+x are x=-4,x=3