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

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

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

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

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y^{2} - y - 3 = 0

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

Decimal Notation

y = 2.30277 \dots, y = - 1.30277 \dots

Solution steps

y^{2} - y - 3 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

- 5 k^{2} = - 20

0 = - x^{2} + 14 x

x^{2} - 12 x + 43

- 5 x^{2} + 94.8 x

7 b - 3 b^{2} = 5 b^{2}

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