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

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

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

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

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- x^{2} + x = - 10

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

Move

10

to the left side

- 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}

12 - 2 x^{2} = - 20

y, x^{2} + x y - y^{2} = 4

y, x^{2} y^{2} - 4 x^{2} - 4 y^{2} = 0

x^{2} - 6 x + 10

x^{2} - 36 = 0

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