What are the solutions to the equation -x^2-2x+19=0 ?

The solutions to the equation -x^2-2x+19=0 are x=-1-2sqrt(5),x=2sqrt(5)-1

Find the zeros of -x^2-2x+19=0

The zeros of -x^2-2x+19=0 are x=-1-2sqrt(5),x=2sqrt(5)-1

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- x^{2} - 2 x + 19 = 0

x = - 1 - 25, x = 25 - 1

Solution steps

- x^{2} - 2 x + 19 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 19}{2 ( - 1 )}

(- 2)^{2} - 4 (- 1) \cdot 19 = 45

x_{1, 2} = \frac{- ( - 2 ) \pm 4 5}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 2 ) + 4 5}{2 ( - 1 )}, x_{2} = \frac{- ( - 2 ) - 4 5}{2 ( - 1 )}

x = \frac{- ( - 2 ) + 4 5}{2 ( - 1 )} : - 1 - 25

x = \frac{- ( - 2 ) - 4 5}{2 ( - 1 )} : 25 - 1

The solutions to the quadratic equation are:

x = - 1 - 25, x = 25 - 1

x^{2} + 4 x + 64 = 0

(7 - x) (- 3 - x) + 26 = 0

3 x^{2} - 6 x + 5 = 0

2 x^{2} - 28 = x (x - 3)

2 x^{2} + 7 x - 3

What are the solutions to the equation -x^2-2x+19=0 ?
The solutions to the equation -x^2-2x+19=0 are x=-1-2sqrt(5),x=2sqrt(5)-1
Find the zeros of -x^2-2x+19=0
The zeros of -x^2-2x+19=0 are x=-1-2sqrt(5),x=2sqrt(5)-1