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

The solutions to the equation y^2+2y=3 are y=1,y=-3

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

The zeros of y^2+2y=3 are y=1,y=-3

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y^{2} + 2 y = 3

y = 1, y = - 3

Solution steps

y^{2} + 2 y = 3

Move

3

to the left side

y^{2} + 2 y - 3 = 0

Solve with the quadratic formula

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

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

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

Separate the solutions

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

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

y = \frac{- 2 - 4}{2 \cdot 1} : - 3

The solutions to the quadratic equation are:

y = 1, y = - 3

x^{2} + 5 x + 6 = 0

y^{2} + 49 = 0

x^{2} + 5 x + 4 = 0

2 x^{2} - m x + (m + 6) = 0

x, 3 x^{2} - 4 x y - 4 y^{2} + 16 x + 16 y - 12 = 0

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