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

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

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

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

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

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

Decimal Notation

q = - 1.71922 \dots, q = - 3.78077 \dots

Solution steps

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

Solve with the quadratic formula

q_{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

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

Separate the solutions

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

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

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

The solutions to the quadratic equation are:

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

3 (x + 1)^{2} - 1 = 11

y, x = (y - 5)^{2}

- 2 (6 x^{2} + 24) = 24

6 - 2 x + 3 x^{2} = 0

x, x^{2} - z^{2} = 5 y

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