What are the solutions to the equation 3y^2+5y=12 ?

The solutions to the equation 3y^2+5y=12 are y= 4/3 ,y=-3

Find the zeros of 3y^2+5y=12

The zeros of 3y^2+5y=12 are y= 4/3 ,y=-3

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3 y^{2} + 5 y = 12

y = \frac{4}{3}, y = - 3

Decimal Notation

y = 1.33333 \dots, y = - 3

Solution steps

3 y^{2} + 5 y = 12

Move

12

to the left side

3 y^{2} + 5 y - 12 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 5 \pm 5 ^{2} - 4 \cdot 3 ( - 12 )}{2 \cdot 3}

5^{2} - 4 \cdot 3 (- 12) = 13

y_{1, 2} = \frac{- 5 \pm 13}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 5 + 13}{2 \cdot 3}, y_{2} = \frac{- 5 - 13}{2 \cdot 3}

y = \frac{- 5 + 13}{2 \cdot 3} : \frac{4}{3}

y = \frac{- 5 - 13}{2 \cdot 3} : - 3

The solutions to the quadratic equation are:

y = \frac{4}{3}, y = - 3

1 2^{\frac{1}{8}} \cdot 1 2^{\frac{5}{6}}

(x + 6) (4 x - 7) = 0

∣ x - 9 ∣ = - 3

3 x^{2} - 4 x = 4

5 \cdot x = 10

What are the solutions to the equation 3y^2+5y=12 ?
The solutions to the equation 3y^2+5y=12 are y= 4/3 ,y=-3
Find the zeros of 3y^2+5y=12
The zeros of 3y^2+5y=12 are y= 4/3 ,y=-3