What are the solutions to the equation 2x^2+11x+13=0 ?

The solutions to the equation 2x^2+11x+13=0 are x=(-11+sqrt(17))/4 ,x=(-11-sqrt(17))/4

Find the zeros of 2x^2+11x+13=0

The zeros of 2x^2+11x+13=0 are x=(-11+sqrt(17))/4 ,x=(-11-sqrt(17))/4

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2 x^{2} + 11 x + 13 = 0

x = \frac{- 11 + 17}{4}, x = \frac{- 11 - 17}{4}

Decimal Notation

x = - 1.71922 \dots, x = - 3.78077 \dots

Solution steps

2 x^{2} + 11 x + 13 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 11 \pm 1 1 ^{2} - 4 \cdot 2 \cdot 13}{2 \cdot 2}

1 1^{2} - 4 \cdot 2 \cdot 13 = 17

x_{1, 2} = \frac{- 11 \pm 17}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 11 + 17}{2 \cdot 2}, x_{2} = \frac{- 11 - 17}{2 \cdot 2}

x = \frac{- 11 + 17}{2 \cdot 2} : \frac{- 11 + 17}{4}

x = \frac{- 11 - 17}{2 \cdot 2} : \frac{- 11 - 17}{4}

The solutions to the quadratic equation are:

x = \frac{- 11 + 17}{4}, x = \frac{- 11 - 17}{4}

2 x^{2} + 3 x = 44

2 x^{2} + 3 x = 11

x^{2} + 8^{2} = 1 3^{2}

t, L {2 t^{2}} = \frac{4}{s ^{3}}

λ, λ^{2} - 2 α λ + λ^{2} - 1 = 0

What are the solutions to the equation 2x^2+11x+13=0 ?
The solutions to the equation 2x^2+11x+13=0 are x=(-11+sqrt(17))/4 ,x=(-11-sqrt(17))/4
Find the zeros of 2x^2+11x+13=0
The zeros of 2x^2+11x+13=0 are x=(-11+sqrt(17))/4 ,x=(-11-sqrt(17))/4