What are the solutions to the equation 0=-x^2-x+3 ?

The solutions to the equation 0=-x^2-x+3 are x=-(1+sqrt(13))/2 ,x=(sqrt(13)-1)/2

Find the zeros of 0=-x^2-x+3

The zeros of 0=-x^2-x+3 are x=-(1+sqrt(13))/2 ,x=(sqrt(13)-1)/2

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0 = - x^{2} - x + 3

x = - \frac{1 + 13}{2}, x = \frac{13 - 1}{2}

Decimal Notation

x = - 2.30277 \dots, x = 1.30277 \dots

Solution steps

0 = - x^{2} - x + 3

Switch sides

- x^{2} - x + 3 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 1 ) \pm ( - 1 ) ^{2} - 4 ( - 1 ) \cdot 3}{2 ( - 1 )}

(- 1)^{2} - 4 (- 1) \cdot 3 = 13

x_{1, 2} = \frac{- ( - 1 ) \pm 13}{2 ( - 1 )}

Separate the solutions

x_{1} = \frac{- ( - 1 ) + 13}{2 ( - 1 )}, x_{2} = \frac{- ( - 1 ) - 13}{2 ( - 1 )}

x = \frac{- ( - 1 ) + 13}{2 ( - 1 )} : - \frac{1 + 13}{2}

x = \frac{- ( - 1 ) - 13}{2 ( - 1 )} : \frac{13 - 1}{2}

The solutions to the quadratic equation are:

x = - \frac{1 + 13}{2}, x = \frac{13 - 1}{2}

f - 1 (- 2) = f (f - 1 (- 2))

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

x^{2} = 784

x = 4 - x^{2}

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

What are the solutions to the equation 0=-x^2-x+3 ?
The solutions to the equation 0=-x^2-x+3 are x=-(1+sqrt(13))/2 ,x=(sqrt(13)-1)/2
Find the zeros of 0=-x^2-x+3
The zeros of 0=-x^2-x+3 are x=-(1+sqrt(13))/2 ,x=(sqrt(13)-1)/2