What are the solutions to the equation -x^2+8x-4=0 ?

The solutions to the equation -x^2+8x-4=0 are x=2(2-sqrt(3)),x=2(2+sqrt(3))

Find the zeros of -x^2+8x-4=0

The zeros of -x^2+8x-4=0 are x=2(2-sqrt(3)),x=2(2+sqrt(3))

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- x^{2} + 8 x - 4 = 0

x = 2 (2 - 3), x = 2 (2 + 3)

Decimal Notation

x = 0.53589 \dots, x = 7.46410 \dots

Solution steps

- x^{2} + 8 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 8 \pm 8 ^{2} - 4 ( - 1 ) ( - 4 )}{2 ( - 1 )}

8^{2} - 4 (- 1) (- 4) = 43

x_{1, 2} = \frac{- 8 \pm 4 3}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 8 + 4 3}{2 ( - 1 )}, x_{2} = \frac{- 8 - 4 3}{2 ( - 1 )}

x = \frac{- 8 + 4 3}{2 ( - 1 )} : 2 (2 - 3)

x = \frac{- 8 - 4 3}{2 ( - 1 )} : 2 (2 + 3)

The solutions to the quadratic equation are:

x = 2 (2 - 3), x = 2 (2 + 3)

- 3 x^{2} - 48 x + 543 = 0

0.3 \cdot 0.3

64 - x^{2} y^{2}

x - 4 (0) = 8

n^{2} + 10 n + 9

What are the solutions to the equation -x^2+8x-4=0 ?
The solutions to the equation -x^2+8x-4=0 are x=2(2-sqrt(3)),x=2(2+sqrt(3))
Find the zeros of -x^2+8x-4=0
The zeros of -x^2+8x-4=0 are x=2(2-sqrt(3)),x=2(2+sqrt(3))