What are the solutions to the equation 39y^2+48y+9=0 ?

The solutions to the equation 39y^2+48y+9=0 are y=-3/13 ,y=-1

Find the zeros of 39y^2+48y+9=0

The zeros of 39y^2+48y+9=0 are y=-3/13 ,y=-1

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39 y^{2} + 48 y + 9 = 0

y = - \frac{3}{13}, y = - 1

Decimal Notation

y = - 0.23076 \dots, y = - 1

Solution steps

39 y^{2} + 48 y + 9 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 48 \pm 4 8 ^{2} - 4 \cdot 39 \cdot 9}{2 \cdot 39}

4 8^{2} - 4 \cdot 39 \cdot 9 = 30

y_{1, 2} = \frac{- 48 \pm 30}{2 \cdot 39}

Separate the solutions

y_{1} = \frac{- 48 + 30}{2 \cdot 39}, y_{2} = \frac{- 48 - 30}{2 \cdot 39}

y = \frac{- 48 + 30}{2 \cdot 39} : - \frac{3}{13}

y = \frac{- 48 - 30}{2 \cdot 39} : - 1

The solutions to the quadratic equation are:

y = - \frac{3}{13}, y = - 1

s^{2} + 1.33 \cdot s + 4.34 = 0

x (2 x - 1) = 3 x - 5 (x + 2)

x, (2 x - 4)^{2} + 3 = 2 (x^{2} + 3) - 4

λ^{2} - 7 λ + 10 (1) = 0

- 16 x^{2} + 16 x + 192 = 0

What are the solutions to the equation 39y^2+48y+9=0 ?
The solutions to the equation 39y^2+48y+9=0 are y=-3/13 ,y=-1
Find the zeros of 39y^2+48y+9=0
The zeros of 39y^2+48y+9=0 are y=-3/13 ,y=-1