What are the solutions to the equation 2y^2+11y+10=0 ?

The solutions to the equation 2y^2+11y+10=0 are y=(-11+sqrt(41))/4 ,y=(-11-sqrt(41))/4

Find the zeros of 2y^2+11y+10=0

The zeros of 2y^2+11y+10=0 are y=(-11+sqrt(41))/4 ,y=(-11-sqrt(41))/4

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2 y^{2} + 11 y + 10 = 0

y = \frac{- 11 + 41}{4}, y = \frac{- 11 - 41}{4}

Decimal Notation

y = - 1.14921 \dots, y = - 4.35078 \dots

Solution steps

2 y^{2} + 11 y + 10 = 0

Solve with the quadratic formula

y_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 \cdot 10}{2 \cdot 2}

1 1^{2} - 4 \cdot 2 \cdot 10 = 41

y_{1, 2} = \frac{- 11 \pm 41}{2 \cdot 2}

Separate the solutions

y_{1} = \frac{- 11 + 41}{2 \cdot 2}, y_{2} = \frac{- 11 - 41}{2 \cdot 2}

y = \frac{- 11 + 41}{2 \cdot 2} : \frac{- 11 + 41}{4}

y = \frac{- 11 - 41}{2 \cdot 2} : \frac{- 11 - 41}{4}

The solutions to the quadratic equation are:

y = \frac{- 11 + 41}{4}, y = \frac{- 11 - 41}{4}

x^{2} - x + 14 = 0

\frac{1}{9} - 1

ln (x^{2} - 3) \leq ln (x - 1)

121 x^{2} - 81 = 0

\frac{8}{π} (e - 1)

What are the solutions to the equation 2y^2+11y+10=0 ?
The solutions to the equation 2y^2+11y+10=0 are y=(-11+sqrt(41))/4 ,y=(-11-sqrt(41))/4
Find the zeros of 2y^2+11y+10=0
The zeros of 2y^2+11y+10=0 are y=(-11+sqrt(41))/4 ,y=(-11-sqrt(41))/4