What are the solutions to the equation y^2+4y-1=0 ?

The solutions to the equation y^2+4y-1=0 are y=-2+sqrt(5),y=-2-sqrt(5)

Find the zeros of y^2+4y-1=0

The zeros of y^2+4y-1=0 are y=-2+sqrt(5),y=-2-sqrt(5)

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y^{2} + 4 y - 1 = 0

y = - 2 + 5, y = - 2 - 5

Decimal Notation

y = 0.23606 \dots, y = - 4.23606 \dots

Solution steps

y^{2} + 4 y - 1 = 0

Solve with the quadratic formula

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

4^{2} - 4 \cdot 1 \cdot (- 1) = 25

y_{1, 2} = \frac{- 4 \pm 2 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 4 + 2 5}{2 \cdot 1}, y_{2} = \frac{- 4 - 2 5}{2 \cdot 1}

y = \frac{- 4 + 2 5}{2 \cdot 1} : - 2 + 5

y = \frac{- 4 - 2 5}{2 \cdot 1} : - 2 - 5

The solutions to the quadratic equation are:

y = - 2 + 5, y = - 2 - 5

24 x^{2} - 82 x + 63 = 0

2 x^{2} + 21 x - 36 = 0

3 t^{2} - 5 t = 1

5 x^{2} = 3 x - 4

x^{2} + (- \frac{3}{6})^{2} = 1

What are the solutions to the equation y^2+4y-1=0 ?
The solutions to the equation y^2+4y-1=0 are y=-2+sqrt(5),y=-2-sqrt(5)
Find the zeros of y^2+4y-1=0
The zeros of y^2+4y-1=0 are y=-2+sqrt(5),y=-2-sqrt(5)