What are the solutions to the equation -x^2-2x+4=0 ?

The solutions to the equation -x^2-2x+4=0 are x=-1-sqrt(5),x=sqrt(5)-1

Find the zeros of -x^2-2x+4=0

The zeros of -x^2-2x+4=0 are x=-1-sqrt(5),x=sqrt(5)-1

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

x = - 1 - 5, x = 5 - 1

Decimal Notation

x = - 3.23606 \dots, x = 1.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 1 - 5, x = 5 - 1

a^{m} \cdot a^{n}

(7^{3})^{3}

(3 + 2) (3 + 2)

8 q - 3 < 10 q + 7

2 (5 x - 2) = 10 x - 4

What are the solutions to the equation -x^2-2x+4=0 ?
The solutions to the equation -x^2-2x+4=0 are x=-1-sqrt(5),x=sqrt(5)-1
Find the zeros of -x^2-2x+4=0
The zeros of -x^2-2x+4=0 are x=-1-sqrt(5),x=sqrt(5)-1