What are the solutions to the equation -x^2+2x+5=0 ?

The solutions to the equation -x^2+2x+5=0 are x=1-sqrt(6),x=1+sqrt(6)

Find the zeros of -x^2+2x+5=0

The zeros of -x^2+2x+5=0 are x=1-sqrt(6),x=1+sqrt(6)

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

x = 1 - 6, x = 1 + 6

Decimal Notation

x = - 1.44948 \dots, x = 3.44948 \dots

Solution steps

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

Solve with the quadratic formula

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

2^{2} - 4 (- 1) \cdot 5 = 26

x_{1, 2} = \frac{- 2 \pm 2 6}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 2 + 2 6}{2 ( - 1 )}, x_{2} = \frac{- 2 - 2 6}{2 ( - 1 )}

x = \frac{- 2 + 2 6}{2 ( - 1 )} : 1 - 6

x = \frac{- 2 - 2 6}{2 ( - 1 )} : 1 + 6

The solutions to the quadratic equation are:

x = 1 - 6, x = 1 + 6

36 x^{2} + 48 x + 16 = 12

9 x^{2} - 42 x = - 52

2 x - 3 = x^{2}

t, h = \frac{1}{2} g t^{2}

- 3 x^{2} - 12 = 0

What are the solutions to the equation -x^2+2x+5=0 ?
The solutions to the equation -x^2+2x+5=0 are x=1-sqrt(6),x=1+sqrt(6)
Find the zeros of -x^2+2x+5=0
The zeros of -x^2+2x+5=0 are x=1-sqrt(6),x=1+sqrt(6)