What are the solutions to the equation 2y^2=-7y-3 ?

The solutions to the equation 2y^2=-7y-3 are y=-1/2 ,y=-3

Find the zeros of 2y^2=-7y-3

The zeros of 2y^2=-7y-3 are y=-1/2 ,y=-3

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2 y^{2} = - 7 y - 3

y = - \frac{1}{2}, y = - 3

Solution steps

2 y^{2} = - 7 y - 3

Move

3

to the left side

2 y^{2} + 3 = - 7 y

Move

7 y

to the left side

2 y^{2} + 3 + 7 y = 0

Write in the standard form

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

2 y^{2} + 7 y + 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 7 \pm 7 ^{2} - 4 \cdot 2 \cdot 3}{2 \cdot 2}

7^{2} - 4 \cdot 2 \cdot 3 = 5

y_{1, 2} = \frac{- 7 \pm 5}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 7 + 5}{2 \cdot 2}, y_{2} = \frac{- 7 - 5}{2 \cdot 2}

y = \frac{- 7 + 5}{2 \cdot 2} : - \frac{1}{2}

y = \frac{- 7 - 5}{2 \cdot 2} : - 3

The solutions to the quadratic equation are:

y = - \frac{1}{2}, y = - 3

t^{2} + 8 t + 12 = 0

w^{2} = 49

(x - 10)^{2} = - 49

\frac{x}{5} - 1 \leq 1

- 8 = - 4 ∣ c ∣

What are the solutions to the equation 2y^2=-7y-3 ?
The solutions to the equation 2y^2=-7y-3 are y=-1/2 ,y=-3
Find the zeros of 2y^2=-7y-3
The zeros of 2y^2=-7y-3 are y=-1/2 ,y=-3