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

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

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

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

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

y = - 1 + 5, y = - 1 - 5

Decimal Notation

y = 1.23606 \dots, y = - 3.23606 \dots

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

y = - 1 + 5, y = - 1 - 5

5 \leq \frac{2}{9} x + 9 \leq 12

\frac{1}{x} + x > 2

429

1 x > 1

27 x^{3} - 8

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