What are the solutions to the equation x^2+13x+5=0 ?

The solutions to the equation x^2+13x+5=0 are x=(-13+sqrt(149))/2 ,x=(-13-sqrt(149))/2

Find the zeros of x^2+13x+5=0

The zeros of x^2+13x+5=0 are x=(-13+sqrt(149))/2 ,x=(-13-sqrt(149))/2

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x^{2} + 13 x + 5 = 0

x = \frac{- 13 + 149}{2}, x = \frac{- 13 - 149}{2}

Decimal Notation

x = - 0.39672 \dots, x = - 12.60327 \dots

Solution steps

x^{2} + 13 x + 5 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 13 \pm 1 3 ^{2} - 4 \cdot 1 \cdot 5}{2 \cdot 1}

1 3^{2} - 4 \cdot 1 \cdot 5 = 149

x_{1, 2} = \frac{- 13 \pm 149}{2 \cdot 1}

Separate the solutions

x_{1} = \frac{- 13 + 149}{2 \cdot 1}, x_{2} = \frac{- 13 - 149}{2 \cdot 1}

x = \frac{- 13 + 149}{2 \cdot 1} : \frac{- 13 + 149}{2}

x = \frac{- 13 - 149}{2 \cdot 1} : \frac{- 13 - 149}{2}

The solutions to the quadratic equation are:

x = \frac{- 13 + 149}{2}, x = \frac{- 13 - 149}{2}

\frac{1}{6} < \frac{2 x - 11}{12} \leq \frac{2}{3}

- \frac{7}{3} + \frac{4}{5} g > - \frac{11}{3}

\frac{1}{4} (2 x - 1) - x < \frac{x}{6} - \frac{1}{3}

\frac{x}{2} + 1 = 5

- 4 (x - 11) = 16

What are the solutions to the equation x^2+13x+5=0 ?
The solutions to the equation x^2+13x+5=0 are x=(-13+sqrt(149))/2 ,x=(-13-sqrt(149))/2
Find the zeros of x^2+13x+5=0
The zeros of x^2+13x+5=0 are x=(-13+sqrt(149))/2 ,x=(-13-sqrt(149))/2