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

The solutions to the equation y^2+3y=-1 are y=(-3+sqrt(5))/2 ,y=(-3-sqrt(5))/2

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

The zeros of y^2+3y=-1 are y=(-3+sqrt(5))/2 ,y=(-3-sqrt(5))/2

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y^{2} + 3 y = - 1

y = \frac{- 3 + 5}{2}, y = \frac{- 3 - 5}{2}

Decimal Notation

y = - 0.38196 \dots, y = - 2.61803 \dots

Solution steps

y^{2} + 3 y = - 1

Move

1

to the left side

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 = 5

y_{1, 2} = \frac{- 3 \pm 5}{2 \cdot 1}

Separate the solutions

y_{1} = \frac{- 3 + 5}{2 \cdot 1}, y_{2} = \frac{- 3 - 5}{2 \cdot 1}

y = \frac{- 3 + 5}{2 \cdot 1} : \frac{- 3 + 5}{2}

y = \frac{- 3 - 5}{2 \cdot 1} : \frac{- 3 - 5}{2}

The solutions to the quadratic equation are:

y = \frac{- 3 + 5}{2}, y = \frac{- 3 - 5}{2}

1 + 1 = 3

\frac{- 1}{3 x + 5} \leq 0

∣ x - 3 ∣ < 11

z^{2} + 4 z = - 3

340

What are the solutions to the equation y^2+3y=-1 ?
The solutions to the equation y^2+3y=-1 are y=(-3+sqrt(5))/2 ,y=(-3-sqrt(5))/2
Find the zeros of y^2+3y=-1
The zeros of y^2+3y=-1 are y=(-3+sqrt(5))/2 ,y=(-3-sqrt(5))/2