What are the solutions to the equation 3y^2+4y=15 ?

The solutions to the equation 3y^2+4y=15 are y= 5/3 ,y=-3

Find the zeros of 3y^2+4y=15

The zeros of 3y^2+4y=15 are y= 5/3 ,y=-3

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3 y^{2} + 4 y = 15

y = \frac{5}{3}, y = - 3

Decimal Notation

y = 1.66666 \dots, y = - 3

Solution steps

3 y^{2} + 4 y = 15

Move

15

to the left side

3 y^{2} + 4 y - 15 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 4 \pm 4 ^{2} - 4 \cdot 3 ( - 15 )}{2 \cdot 3}

4^{2} - 4 \cdot 3 (- 15) = 14

y_{1, 2} = \frac{- 4 \pm 14}{2 \cdot 3}

Separate the solutions

y_{1} = \frac{- 4 + 14}{2 \cdot 3}, y_{2} = \frac{- 4 - 14}{2 \cdot 3}

y = \frac{- 4 + 14}{2 \cdot 3} : \frac{5}{3}

y = \frac{- 4 - 14}{2 \cdot 3} : - 3

The solutions to the quadratic equation are:

y = \frac{5}{3}, y = - 3

3 + 2 x - x^{2} = 0

x^{2} + 10 x - 50 = 0

2^{2} + 3^{2} = c^{2}

1 1^{2} + x^{2} = 6 1^{2}

x^{2} + 10 x - 57 = 4

What are the solutions to the equation 3y^2+4y=15 ?
The solutions to the equation 3y^2+4y=15 are y= 5/3 ,y=-3
Find the zeros of 3y^2+4y=15
The zeros of 3y^2+4y=15 are y= 5/3 ,y=-3