What are the solutions to the equation 3y^2+7y+4=0 ?

The solutions to the equation 3y^2+7y+4=0 are y=-1,y=-4/3

Find the zeros of 3y^2+7y+4=0

The zeros of 3y^2+7y+4=0 are y=-1,y=-4/3

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3 y^{2} + 7 y + 4 = 0

y = - 1, y = - \frac{4}{3}

Decimal Notation

y = - 1, y = - 1.33333 \dots

Solution steps

3 y^{2} + 7 y + 4 = 0

Solve with the quadratic formula

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

7^{2} - 4 \cdot 3 \cdot 4 = 1

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

Separate the solutions

y_{1} = \frac{- 7 + 1}{2 \cdot 3}, y_{2} = \frac{- 7 - 1}{2 \cdot 3}

y = \frac{- 7 + 1}{2 \cdot 3} : - 1

y = \frac{- 7 - 1}{2 \cdot 3} : - \frac{4}{3}

The solutions to the quadratic equation are:

y = - 1, y = - \frac{4}{3}

3 x^{2} - 2 x = - 5

x^{2} - 10 x + 21 = 0

x^{2} - 10 x + 24 = 0

r^{2} - 6 r + 5 = 0

x, Lx - x^{2} - \frac{1}{6} L^{2} = 0

What are the solutions to the equation 3y^2+7y+4=0 ?
The solutions to the equation 3y^2+7y+4=0 are y=-1,y=-4/3
Find the zeros of 3y^2+7y+4=0
The zeros of 3y^2+7y+4=0 are y=-1,y=-4/3