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

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

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

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

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

x = 2 - 3, x = 2 + 3

Decimal Notation

x = 0.26794 \dots, x = 3.73205 \dots

Solution steps

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

Switch sides

- 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

(x + 4) (x - 4) - (x - 2)^{2} = x (x - 5)

x = - x^{2} + 1000 x

- x^{2} + 3 x + a = 0

0 = 12 x^{2} - 24 x - 40

- 6 x^{2} + 72 x - 192 = 0

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