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=1+sqrt(5)

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

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

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

x = 1 - 5, x = 1 + 5

Decimal Notation

x = - 1.23606 \dots, x = 3.23606 \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 )} : 1 + 5

The solutions to the quadratic equation are:

x = 1 - 5, x = 1 + 5

\frac{1}{3} - 4

(2 - 3 i)^{2}

- 3 t - 2 = - 2 t + 1

- 3 x^{2} - 16 x - 16 = 0

\frac{( 5 ^{3} ) ^{9}}{5 ^{12}}

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