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

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

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

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

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

y = - 3, y = 1

Solution steps

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

Solve with the quadratic formula

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

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

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

Separate the solutions

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

y = \frac{- ( - 2 ) + 4}{2 ( - 1 )} : - 3

y = \frac{- ( - 2 ) - 4}{2 ( - 1 )} : 1

The solutions to the quadratic equation are:

y = - 3, y = 1

a x^{2} - 5 x^{2} + 2 x = a + 7

x, x^{2} + y^{2} + 3 x + 2 y + 5 = 0

Z^{2} - 5 = - 4

2 (x^{2} + 1) = - 2 x

n^{2} - 4 n - 21 = 0

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