What are the solutions to the equation (-7-x)(7-x)+24=0 ?

The solutions to the equation (-7-x)(7-x)+24=0 are x=5,x=-5

Find the zeros of (-7-x)(7-x)+24=0

The zeros of (-7-x)(7-x)+24=0 are x=5,x=-5

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(- 7 - x) (7 - x) + 24 = 0

x = 5, x = - 5

Solution steps

(- 7 - x) (7 - x) + 24 = 0

Expand

(- 7 - x) (7 - x) + 24 : x^{2} - 25

x^{2} - 25 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 0 \pm 0 ^{2} - 4 \cdot 1 \cdot ( - 25 )}{2 \cdot 1}

0^{2} - 4 \cdot 1 \cdot (- 25) = 10

x_{1, 2} = \frac{- 0 \pm 10}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 0 + 10}{2 \cdot 1}, x_{2} = \frac{- 0 - 10}{2 \cdot 1}

x = \frac{- 0 + 10}{2 \cdot 1} : 5

x = \frac{- 0 - 10}{2 \cdot 1} : - 5

The solutions to the quadratic equation are:

x = 5, x = - 5

0.2 x^{2} - 2.2 x - 3.8 = 0

x^{2} + 6 = 70

9 x^{2} + 4 x - 3 = 0

2 x^{2} + 4 x + 7

t, x = v t + \frac{1}{2} a t^{2}

What are the solutions to the equation (-7-x)(7-x)+24=0 ?
The solutions to the equation (-7-x)(7-x)+24=0 are x=5,x=-5
Find the zeros of (-7-x)(7-x)+24=0
The zeros of (-7-x)(7-x)+24=0 are x=5,x=-5