What are the solutions to the equation -x^2+x+10=0 ?

The solutions to the equation -x^2+x+10=0 are x=-(-1+sqrt(41))/2 ,x=(1+sqrt(41))/2

Find the zeros of -x^2+x+10=0

The zeros of -x^2+x+10=0 are x=-(-1+sqrt(41))/2 ,x=(1+sqrt(41))/2

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- x^{2} + x + 10 = 0

x = - \frac{- 1 + 41}{2}, x = \frac{1 + 41}{2}

Decimal Notation

x = - 2.70156 \dots, x = 3.70156 \dots

Solution steps

- x^{2} + x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 1 \pm 1 ^{2} - 4 ( - 1 ) \cdot 10}{2 ( - 1 )}

1^{2} - 4 (- 1) \cdot 10 = 41

x_{1, 2} = \frac{- 1 \pm 41}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 41}{2 ( - 1 )}, x_{2} = \frac{- 1 - 41}{2 ( - 1 )}

x = \frac{- 1 + 41}{2 ( - 1 )} : - \frac{- 1 + 41}{2}

x = \frac{- 1 - 41}{2 ( - 1 )} : \frac{1 + 41}{2}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 41}{2}, x = \frac{1 + 41}{2}

4 x^{2} - 15 x + 9 = 0

\frac{3}{280} + \frac{1}{150} + \frac{11}{225}

5.3 - (- 3.7 c) = 13.81

63 n^{3} + 54 n^{2} - 105 n - 90

\frac{x - 3}{x + 1} < 0

What are the solutions to the equation -x^2+x+10=0 ?
The solutions to the equation -x^2+x+10=0 are x=-(-1+sqrt(41))/2 ,x=(1+sqrt(41))/2
Find the zeros of -x^2+x+10=0
The zeros of -x^2+x+10=0 are x=-(-1+sqrt(41))/2 ,x=(1+sqrt(41))/2