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

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

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

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

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

x = - 7, x = 5

Solution steps

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

Solve with the quadratic formula

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

(- 2)^{2} - 4 (- 1) \cdot 35 = 12

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

x = - 7, x = 5

8 a^{2} + 12 a

8 (3)

x^{12} - y^{12}

\frac{6 n}{5} + \frac{1}{2} = \frac{4 n - 3}{10}

6 t^{2} + 5 = 2 t^{2} + 1

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