What are the solutions to the equation 7+2x-x^2=0 ?

The solutions to the equation 7+2x-x^2=0 are x=1-2sqrt(2),x=1+2sqrt(2)

Find the zeros of 7+2x-x^2=0

The zeros of 7+2x-x^2=0 are x=1-2sqrt(2),x=1+2sqrt(2)

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7 + 2 x - x^{2} = 0

x = 1 - 22, x = 1 + 22

Solution steps

7 + 2 x - x^{2} = 0

Write in the standard form

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

- x^{2} + 2 x + 7 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 ( - 1 ) \cdot 7}{2 ( - 1 )}

2^{2} - 4 (- 1) \cdot 7 = 42

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

Separate the solutions

x_{1} = \frac{- 2 + 4 2}{2 ( - 1 )}, x_{2} = \frac{- 2 - 4 2}{2 ( - 1 )}

x = \frac{- 2 + 4 2}{2 ( - 1 )} : 1 - 22

x = \frac{- 2 - 4 2}{2 ( - 1 )} : 1 + 22

The solutions to the quadratic equation are:

x = 1 - 22, x = 1 + 22

\frac{( 2 a ^{5} b ) ^{2}}{24 b ^{6}}

- 3 x^{2} - 15 + 12 x = 0

(x - 8) (x - 2) (x + 4) \leq 0

n^{2} - 81

2 x^{2} + 11 x - 4 = 2

What are the solutions to the equation 7+2x-x^2=0 ?
The solutions to the equation 7+2x-x^2=0 are x=1-2sqrt(2),x=1+2sqrt(2)
Find the zeros of 7+2x-x^2=0
The zeros of 7+2x-x^2=0 are x=1-2sqrt(2),x=1+2sqrt(2)