What are the solutions to the equation y^2+y-4=0 ?

The solutions to the equation y^2+y-4=0 are y=(-1+sqrt(17))/2 ,y=(-1-sqrt(17))/2

Find the zeros of y^2+y-4=0

The zeros of y^2+y-4=0 are y=(-1+sqrt(17))/2 ,y=(-1-sqrt(17))/2

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y^{2} + y - 4 = 0

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

Decimal Notation

y = 1.56155 \dots, y = - 2.56155 \dots

Solution steps

y^{2} + y - 4 = 0

Solve with the quadratic formula

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

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

y_{1, 2} = \frac{- 1 \pm 17}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 1 + 17}{2 \cdot 1}, y_{2} = \frac{- 1 - 17}{2 \cdot 1}

y = \frac{- 1 + 17}{2 \cdot 1} : \frac{- 1 + 17}{2}

y = \frac{- 1 - 17}{2 \cdot 1} : \frac{- 1 - 17}{2}

The solutions to the quadratic equation are:

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

36 p^{2} + 12 p + 1 = 0

9 (5 - 5 x)^{2} + 10 = 11

x^{2} + 76 = 18 x

x^{2} - 14 x + 49 = 9

3 x^{2} - 12 x + 6 = 0

What are the solutions to the equation y^2+y-4=0 ?
The solutions to the equation y^2+y-4=0 are y=(-1+sqrt(17))/2 ,y=(-1-sqrt(17))/2
Find the zeros of y^2+y-4=0
The zeros of y^2+y-4=0 are y=(-1+sqrt(17))/2 ,y=(-1-sqrt(17))/2