What are the solutions to the equation -x^2+169x-28=0 ?

The solutions to the equation -x^2+169x-28=0 are x=(169-3sqrt(3161))/2 ,x=(169+3sqrt(3161))/2

Find the zeros of -x^2+169x-28=0

The zeros of -x^2+169x-28=0 are x=(169-3sqrt(3161))/2 ,x=(169+3sqrt(3161))/2

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- x^{2} + 169 x - 28 = 0

x = \frac{169 - 3 3161}{2}, x = \frac{169 + 3 3161}{2}

Decimal Notation

x = 0.16584 \dots, x = 168.83415 \dots

Solution steps

- x^{2} + 169 x - 28 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 169 \pm 16 9 ^{2} - 4 ( - 1 ) ( - 28 )}{2 ( - 1 )}

16 9^{2} - 4 (- 1) (- 28) = 33161

x_{1, 2} = \frac{- 169 \pm 3 3161}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 169 + 3 3161}{2 ( - 1 )}, x_{2} = \frac{- 169 - 3 3161}{2 ( - 1 )}

x = \frac{- 169 + 3 3161}{2 ( - 1 )} : \frac{169 - 3 3161}{2}

x = \frac{- 169 - 3 3161}{2 ( - 1 )} : \frac{169 + 3 3161}{2}

The solutions to the quadratic equation are:

x = \frac{169 - 3 3161}{2}, x = \frac{169 + 3 3161}{2}

x, \frac{x}{y} = \frac{x ^{2}}{y ^{2}}

6 x - 1 \cdot 6 x \cdot 6 x + 2 = 11296

\frac{4}{3} p^{2} - 1 = 0

3 x^{2} - 22 x + 36 = 0

x^{2} + \frac{8 x}{3} - 1 = 0

What are the solutions to the equation -x^2+169x-28=0 ?
The solutions to the equation -x^2+169x-28=0 are x=(169-3sqrt(3161))/2 ,x=(169+3sqrt(3161))/2
Find the zeros of -x^2+169x-28=0
The zeros of -x^2+169x-28=0 are x=(169-3sqrt(3161))/2 ,x=(169+3sqrt(3161))/2