What are the solutions to the equation -x^2+4x-1=0 ?

The solutions to the equation -x^2+4x-1=0 are x=2-sqrt(3),x=2+sqrt(3)

Find the zeros of -x^2+4x-1=0

The zeros of -x^2+4x-1=0 are x=2-sqrt(3),x=2+sqrt(3)

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- x^{2} + 4 x - 1 = 0

x = 2 - 3, x = 2 + 3

Decimal Notation

x = 0.26794 \dots, x = 3.73205 \dots

Solution steps

- x^{2} + 4 x - 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 (- 1) (- 1) = 23

x_{1, 2} = \frac{- 4 \pm 2 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 4 + 2 3}{2 ( - 1 )}, x_{2} = \frac{- 4 - 2 3}{2 ( - 1 )}

x = \frac{- 4 + 2 3}{2 ( - 1 )} : 2 - 3

x = \frac{- 4 - 2 3}{2 ( - 1 )} : 2 + 3

The solutions to the quadratic equation are:

x = 2 - 3, x = 2 + 3

3 t^{2} - 12 t + 11 = 0

x^{2} - 3 x + x y - 3 y

(9 - 8 i) - (- 6 - i)

2 x^{2} - 3 x y

x^{5} - 4 x^{4} - x^{3} + 10 x^{2} + 2 x - 4

What are the solutions to the equation -x^2+4x-1=0 ?
The solutions to the equation -x^2+4x-1=0 are x=2-sqrt(3),x=2+sqrt(3)
Find the zeros of -x^2+4x-1=0
The zeros of -x^2+4x-1=0 are x=2-sqrt(3),x=2+sqrt(3)