What are the solutions to the equation -y^2-2y+18=0 ?

The solutions to the equation -y^2-2y+18=0 are y=-1-sqrt(19),y=sqrt(19)-1

Find the zeros of -y^2-2y+18=0

The zeros of -y^2-2y+18=0 are y=-1-sqrt(19),y=sqrt(19)-1

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- y^{2} - 2 y + 18 = 0

y = - 1 - 19, y = 19 - 1

Solution steps

- y^{2} - 2 y + 18 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 2 ) \pm ( - 2 ) ^{2} - 4 ( - 1 ) \cdot 18}{2 ( - 1 )}

(- 2)^{2} - 4 (- 1) \cdot 18 = 219

y_{1, 2} = \frac{- ( - 2 ) \pm 2 19}{2 ( - 1 )}

Separate the solutions

y_{1} = \frac{- ( - 2 ) + 2 19}{2 ( - 1 )}, y_{2} = \frac{- ( - 2 ) - 2 19}{2 ( - 1 )}

y = \frac{- ( - 2 ) + 2 19}{2 ( - 1 )} : - 1 - 19

y = \frac{- ( - 2 ) - 2 19}{2 ( - 1 )} : 19 - 1

The solutions to the quadratic equation are:

y = - 1 - 19, y = 19 - 1

11 \cdot (\frac{7}{4})

3 x^{2} - 2 x - 13 = - 6

\frac{3}{4} x - \frac{1}{2} = 4

4 y < 32

∣ x + 6 ∣ = 2 x

What are the solutions to the equation -y^2-2y+18=0 ?
The solutions to the equation -y^2-2y+18=0 are y=-1-sqrt(19),y=sqrt(19)-1
Find the zeros of -y^2-2y+18=0
The zeros of -y^2-2y+18=0 are y=-1-sqrt(19),y=sqrt(19)-1