What are the solutions to the equation 3y-y^2+1=0 ?

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

Find the zeros of 3y-y^2+1=0

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

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3 y - y^{2} + 1 = 0

y = - \frac{- 3 + 13}{2}, y = \frac{3 + 13}{2}

Decimal Notation

y = - 0.30277 \dots, y = 3.30277 \dots

Solution steps

3 y - y^{2} + 1 = 0

Write in the standard form

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

- y^{2} + 3 y + 1 = 0

Solve with the quadratic formula

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

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

y_{1, 2} = \frac{- 3 \pm 13}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- 3 + 13}{2 ( - 1 )}, y_{2} = \frac{- 3 - 13}{2 ( - 1 )}

y = \frac{- 3 + 13}{2 ( - 1 )} : - \frac{- 3 + 13}{2}

y = \frac{- 3 - 13}{2 ( - 1 )} : \frac{3 + 13}{2}

The solutions to the quadratic equation are:

y = - \frac{- 3 + 13}{2}, y = \frac{3 + 13}{2}

w (w + 3) = 154

x (x - 1) = x (7 - x)

3 x^{2} + 30 x + 150 = 0

3 x^{2} + a x + 12 = 0

5 x^{2} - 10 x + 9 = 0

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