What are the solutions to the equation 13x^2+1=-10x ?

The solutions to the equation 13x^2+1=-10x are x=(-5+2sqrt(3))/(13),x=-(5+2sqrt(3))/(13)

Find the zeros of 13x^2+1=-10x

The zeros of 13x^2+1=-10x are x=(-5+2sqrt(3))/(13),x=-(5+2sqrt(3))/(13)

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13 x^{2} + 1 = - 10 x

x = \frac{- 5 + 2 3}{13}, x = - \frac{5 + 2 3}{13}

Solution steps

13 x^{2} + 1 = - 10 x

Move

10 x

to the left side

13 x^{2} + 1 + 10 x = 0

Write in the standard form

a x^{2} + b x + c = 0

13 x^{2} + 10 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 10 \pm 1 0 ^{2} - 4 \cdot 13 \cdot 1}{2 \cdot 13}

1 0^{2} - 4 \cdot 13 \cdot 1 = 43

x_{1, 2} = \frac{- 10 \pm 4 3}{2 \cdot 13}

Separate the solutions

x_{1} = \frac{- 10 + 4 3}{2 \cdot 13}, x_{2} = \frac{- 10 - 4 3}{2 \cdot 13}

x = \frac{- 10 + 4 3}{2 \cdot 13} : \frac{- 5 + 2 3}{13}

x = \frac{- 10 - 4 3}{2 \cdot 13} : - \frac{5 + 2 3}{13}

The solutions to the quadratic equation are:

x = \frac{- 5 + 2 3}{13}, x = - \frac{5 + 2 3}{13}

\frac{4}{3} - 1

x^{6} - b^{9}

(12 m)^{\frac{1}{3}}

\frac{3}{8} + \frac{5}{6}

(x + 2)^{2} = - 4

What are the solutions to the equation 13x^2+1=-10x ?
The solutions to the equation 13x^2+1=-10x are x=(-5+2sqrt(3))/(13),x=-(5+2sqrt(3))/(13)
Find the zeros of 13x^2+1=-10x
The zeros of 13x^2+1=-10x are x=(-5+2sqrt(3))/(13),x=-(5+2sqrt(3))/(13)