What are the solutions to the equation 4-x^2=-x ?

The solutions to the equation 4-x^2=-x are x=-(-1+sqrt(17))/2 ,x=(1+sqrt(17))/2

Find the zeros of 4-x^2=-x

The zeros of 4-x^2=-x are x=-(-1+sqrt(17))/2 ,x=(1+sqrt(17))/2

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4 - x^{2} = - x

x = - \frac{- 1 + 17}{2}, x = \frac{1 + 17}{2}

Decimal Notation

x = - 1.56155 \dots, x = 2.56155 \dots

Solution steps

4 - x^{2} = - x

Move

x

to the left side

4 - x^{2} + x = 0

Write in the standard form

a x^{2} + b x + c = 0

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

Solve with the quadratic formula

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

1^{2} - 4 (- 1) \cdot 4 = 17

x_{1, 2} = \frac{- 1 \pm 17}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- 1 + 17}{2 ( - 1 )}, x_{2} = \frac{- 1 - 17}{2 ( - 1 )}

x = \frac{- 1 + 17}{2 ( - 1 )} : - \frac{- 1 + 17}{2}

x = \frac{- 1 - 17}{2 ( - 1 )} : \frac{1 + 17}{2}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 17}{2}, x = \frac{1 + 17}{2}

4 q^{2} + 9 q + 4 = 0

s^{2} + 2 s + 13 = 0

x, x^{2} + 8 x - 20 = 0

x^{2} + 6 x - 40 = 0

y, \frac{x ^{2}}{81} + \frac{y ^{2}}{144} = 1

What are the solutions to the equation 4-x^2=-x ?
The solutions to the equation 4-x^2=-x are x=-(-1+sqrt(17))/2 ,x=(1+sqrt(17))/2
Find the zeros of 4-x^2=-x
The zeros of 4-x^2=-x are x=-(-1+sqrt(17))/2 ,x=(1+sqrt(17))/2