What are the solutions to the equation 3x^2+1-7x=13x^2 ?

The solutions to the equation 3x^2+1-7x=13x^2 are x=-(7+sqrt(89))/(20),x=(sqrt(89)-7)/(20)

Find the zeros of 3x^2+1-7x=13x^2

The zeros of 3x^2+1-7x=13x^2 are x=-(7+sqrt(89))/(20),x=(sqrt(89)-7)/(20)

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3 x^{2} + 1 - 7 x = 13 x^{2}

x = - \frac{7 + 89}{20}, x = \frac{89 - 7}{20}

Decimal Notation

x = - 0.82169 \dots, x = 0.12169 \dots

Solution steps

3 x^{2} + 1 - 7 x = 13 x^{2}

Move

13 x^{2}

to the left side

- 10 x^{2} - 7 x + 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 7 ) \pm ( - 7 ) ^{2} - 4 ( - 10 ) \cdot 1}{2 ( - 10 )}

(- 7)^{2} - 4 (- 10) \cdot 1 = 89

x_{1, 2} = \frac{- ( - 7 ) \pm 89}{2 ( - 10 )}

Separate the solutions

x_{1} = \frac{- ( - 7 ) + 89}{2 ( - 10 )}, x_{2} = \frac{- ( - 7 ) - 89}{2 ( - 10 )}

x = \frac{- ( - 7 ) + 89}{2 ( - 10 )} : - \frac{7 + 89}{20}

x = \frac{- ( - 7 ) - 89}{2 ( - 10 )} : \frac{89 - 7}{20}

The solutions to the quadratic equation are:

x = - \frac{7 + 89}{20}, x = \frac{89 - 7}{20}

(e^{x}) (e^{x})

15 y^{2} + 14 y - 16

2 y^{2} - 1 = 0

4 - 2 ∣ x - 3 ∣ > - 2

- 2 x - 2 = x - 14

What are the solutions to the equation 3x^2+1-7x=13x^2 ?
The solutions to the equation 3x^2+1-7x=13x^2 are x=-(7+sqrt(89))/(20),x=(sqrt(89)-7)/(20)
Find the zeros of 3x^2+1-7x=13x^2
The zeros of 3x^2+1-7x=13x^2 are x=-(7+sqrt(89))/(20),x=(sqrt(89)-7)/(20)