What are the solutions to the equation -x^2+168x-5292=0 ?

The solutions to the equation -x^2+168x-5292=0 are x=42,x=126

Find the zeros of -x^2+168x-5292=0

The zeros of -x^2+168x-5292=0 are x=42,x=126

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- x^{2} + 168 x - 5292 = 0

x = 42, x = 126

Solution steps

- x^{2} + 168 x - 5292 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 168 \pm 16 8 ^{2} - 4 ( - 1 ) ( - 5292 )}{2 ( - 1 )}

16 8^{2} - 4 (- 1) (- 5292) = 84

x_{1, 2} = \frac{- 168 \pm 84}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 168 + 84}{2 ( - 1 )}, x_{2} = \frac{- 168 - 84}{2 ( - 1 )}

x = \frac{- 168 + 84}{2 ( - 1 )} : 42

x = \frac{- 168 - 84}{2 ( - 1 )} : 126

The solutions to the quadratic equation are:

x = 42, x = 126

- 6 \cdot \frac{4}{3}

\frac{x - 2}{2} = \frac{3 x + 6}{3}

(\frac{x}{y})^{- 2}

3^{5} \cdot 3^{- 3}

f = 5^{1}

What are the solutions to the equation -x^2+168x-5292=0 ?
The solutions to the equation -x^2+168x-5292=0 are x=42,x=126
Find the zeros of -x^2+168x-5292=0
The zeros of -x^2+168x-5292=0 are x=42,x=126