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

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

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

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

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

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

Decimal Notation

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

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

y = \frac{- 3 - 13}{2 \cdot 1} : \frac{- 3 - 13}{2}

The solutions to the quadratic equation are:

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

z^{2} - 3 z = 54

s^{2} + 10 s + 61

3 x^{2} - 8 x + 4 = 0

3 x^{2} + 7 x = 9

x, x^{2} + y^{2} - 14 x + 2 y + 50 = 0

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