What are the solutions to the equation -4y-2y^2=-16 ?

The solutions to the equation -4y-2y^2=-16 are y=-4,y=2

Find the zeros of -4y-2y^2=-16

The zeros of -4y-2y^2=-16 are y=-4,y=2

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- 4 y - 2 y^{2} = - 16

y = - 4, y = 2

Solution steps

- 4 y - 2 y^{2} = - 16

Move

16

to the left side

- 4 y - 2 y^{2} + 16 = 0

Write in the standard form

a x^{2} + b x + c = 0

- 2 y^{2} - 4 y + 16 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 ( - 2 ) \cdot 16}{2 ( - 2 )}

(- 4)^{2} - 4 (- 2) \cdot 16 = 12

y_{1, 2} = \frac{- ( - 4 ) \pm 12}{2 ( - 2 )}

Separate the solutions

y_{1} = \frac{- ( - 4 ) + 12}{2 ( - 2 )}, y_{2} = \frac{- ( - 4 ) - 12}{2 ( - 2 )}

y = \frac{- ( - 4 ) + 12}{2 ( - 2 )} : - 4

y = \frac{- ( - 4 ) - 12}{2 ( - 2 )} : 2

The solutions to the quadratic equation are:

y = - 4, y = 2

3 x^{2} = - 75

(3 x + 1) (2 x - 1) = 0

2 x^{2} = 2 x + 1

25 - 70 X + 49 X^{2} = 0

x, \frac{x ^{2}}{25} - \frac{y ^{2}}{16} = 0

What are the solutions to the equation -4y-2y^2=-16 ?
The solutions to the equation -4y-2y^2=-16 are y=-4,y=2
Find the zeros of -4y-2y^2=-16
The zeros of -4y-2y^2=-16 are y=-4,y=2