What are the solutions to the equation 4y^2+11y+3=-1 ?

The solutions to the equation 4y^2+11y+3=-1 are y=(-11+sqrt(57))/8 ,y=(-11-sqrt(57))/8

Find the zeros of 4y^2+11y+3=-1

The zeros of 4y^2+11y+3=-1 are y=(-11+sqrt(57))/8 ,y=(-11-sqrt(57))/8

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4 y^{2} + 11 y + 3 = - 1

y = \frac{- 11 + 57}{8}, y = \frac{- 11 - 57}{8}

Decimal Notation

y = - 0.43127 \dots, y = - 2.31872 \dots

Solution steps

4 y^{2} + 11 y + 3 = - 1

Move

1

to the left side

4 y^{2} + 11 y + 4 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 4 \cdot 4}{2 \cdot 4}

1 1^{2} - 4 \cdot 4 \cdot 4 = 57

y_{1, 2} = \frac{- 11 \pm 57}{2 \cdot 4}

Separate the solutions

y_{1} = \frac{- 11 + 57}{2 \cdot 4}, y_{2} = \frac{- 11 - 57}{2 \cdot 4}

y = \frac{- 11 + 57}{2 \cdot 4} : \frac{- 11 + 57}{8}

y = \frac{- 11 - 57}{2 \cdot 4} : \frac{- 11 - 57}{8}

The solutions to the quadratic equation are:

y = \frac{- 11 + 57}{8}, y = \frac{- 11 - 57}{8}

v, \frac{m v ^{2}}{r} = q v B

5 x^{2} + 14 x = x + 6

x, x^{2} (y^{2} - 4) = 4

x, 4 y^{2} + 10 x^{2} - 2 y + 5 = 0

4 x^{2} + 48 x - 25 = 0

What are the solutions to the equation 4y^2+11y+3=-1 ?
The solutions to the equation 4y^2+11y+3=-1 are y=(-11+sqrt(57))/8 ,y=(-11-sqrt(57))/8
Find the zeros of 4y^2+11y+3=-1
The zeros of 4y^2+11y+3=-1 are y=(-11+sqrt(57))/8 ,y=(-11-sqrt(57))/8