What are the solutions to the equation -x^2+12x-35=0 ?

The solutions to the equation -x^2+12x-35=0 are x=5,x=7

Find the zeros of -x^2+12x-35=0

The zeros of -x^2+12x-35=0 are x=5,x=7

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- x^{2} + 12 x - 35 = 0

x = 5, x = 7

Solution steps

- x^{2} + 12 x - 35 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 1 ) ( - 35 )}{2 ( - 1 )}

1 2^{2} - 4 (- 1) (- 35) = 2

x_{1, 2} = \frac{- 12 \pm 2}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 12 + 2}{2 ( - 1 )}, x_{2} = \frac{- 12 - 2}{2 ( - 1 )}

x = \frac{- 12 + 2}{2 ( - 1 )} : 5

x = \frac{- 12 - 2}{2 ( - 1 )} : 7

The solutions to the quadratic equation are:

x = 5, x = 7

3 (\frac{5}{2}) - 2 y + 2 = 0

\frac{38 !}{26 ! 12 !}

3 \cdot \frac{1}{5}

(x - 4) (x + 7) = 0

x^{2} - 8 \geq 0

What are the solutions to the equation -x^2+12x-35=0 ?
The solutions to the equation -x^2+12x-35=0 are x=5,x=7
Find the zeros of -x^2+12x-35=0
The zeros of -x^2+12x-35=0 are x=5,x=7