What are the solutions to the equation (-5-u)(-5+u)=9 ?

The solutions to the equation (-5-u)(-5+u)=9 are u=-4,u=4

Find the zeros of (-5-u)(-5+u)=9

The zeros of (-5-u)(-5+u)=9 are u=-4,u=4

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(- 5 - u) (- 5 + u) = 9

u = - 4, u = 4

Solution steps

(- 5 - u) (- 5 + u) = 9

Expand

(- 5 - u) (- 5 + u) : 25 - u^{2}

25 - u^{2} = 9

Move

9

to the left side

- u^{2} + 16 = 0

Solve with the quadratic formula

u_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 ( - 1 ) \cdot 16}{2 ( - 1 )}

0^{2} - 4 (- 1) \cdot 16 = 8

u_{1, 2} = \frac{- 0 \pm 8}{2 ( - 1 )}

Separate the solutions

u_{1} = \frac{- 0 + 8}{2 ( - 1 )}, u_{2} = \frac{- 0 - 8}{2 ( - 1 )}

u = \frac{- 0 + 8}{2 ( - 1 )} : - 4

u = \frac{- 0 - 8}{2 ( - 1 )} : 4

The solutions to the quadratic equation are:

u = - 4, u = 4

x, x^{2} + 6 x - 4 = 0

3 = x^{2} + 2 x + 1

40 = x^{2} - 6 x

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

- x^{2} - 5 x - 17 = 0

What are the solutions to the equation (-5-u)(-5+u)=9 ?
The solutions to the equation (-5-u)(-5+u)=9 are u=-4,u=4
Find the zeros of (-5-u)(-5+u)=9
The zeros of (-5-u)(-5+u)=9 are u=-4,u=4