What are the solutions to the equation y^2+2y-3=0 ?

The solutions to the equation y^2+2y-3=0 are y=1,y=-3

Find the zeros of y^2+2y-3=0

The zeros of y^2+2y-3=0 are y=1,y=-3

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

y = 1, y = - 3

Solution steps

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} - 11 x = 24

x, 3 x^{2} + 5 y^{2} - 6 x - 20 y = - 25

2 x^{2} + 144 = - 32 x

x, x^{2} + 4 = 0

- x^{2} + 6 x = (x + 3)^{2} - 9

What are the solutions to the equation y^2+2y-3=0 ?
The solutions to the equation y^2+2y-3=0 are y=1,y=-3
Find the zeros of y^2+2y-3=0
The zeros of y^2+2y-3=0 are y=1,y=-3