What are the solutions to the equation 3y^2+10y=-3 ?

The solutions to the equation 3y^2+10y=-3 are y=-1/3 ,y=-3

Find the zeros of 3y^2+10y=-3

The zeros of 3y^2+10y=-3 are y=-1/3 ,y=-3

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3 y^{2} + 10 y = - 3

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

Decimal Notation

y = - 0.33333 \dots, y = - 3

Solution steps

3 y^{2} + 10 y = - 3

Move

3

to the left side

3 y^{2} + 10 y + 3 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 3 \cdot 3}{2 \cdot 3}

1 0^{2} - 4 \cdot 3 \cdot 3 = 8

y_{1, 2} = \frac{- 10 \pm 8}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 10 + 8}{2 \cdot 3}, y_{2} = \frac{- 10 - 8}{2 \cdot 3}

y = \frac{- 10 + 8}{2 \cdot 3} : - \frac{1}{3}

y = \frac{- 10 - 8}{2 \cdot 3} : - 3

The solutions to the quadratic equation are:

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

- 9 y^{2} + 6 y + 1 = 0

5 a^{2} - 13 a = 0

9 m^{2} - 4 = 20

m^{2} + 13 m = m

4 x^{2} = - 2 x - 1

What are the solutions to the equation 3y^2+10y=-3 ?
The solutions to the equation 3y^2+10y=-3 are y=-1/3 ,y=-3
Find the zeros of 3y^2+10y=-3
The zeros of 3y^2+10y=-3 are y=-1/3 ,y=-3