What are the solutions to the equation 6x^2-62x+99=0 ?

The solutions to the equation 6x^2-62x+99=0 are x=(31+sqrt(367))/6 ,x=(31-sqrt(367))/6

Find the zeros of 6x^2-62x+99=0

The zeros of 6x^2-62x+99=0 are x=(31+sqrt(367))/6 ,x=(31-sqrt(367))/6

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6 x^{2} - 62 x + 99 = 0

x = \frac{31 + 367}{6}, x = \frac{31 - 367}{6}

Decimal Notation

x = 8.35954 \dots, x = 1.97379 \dots

Solution steps

6 x^{2} - 62 x + 99 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 62 ) \pm ( - 62 ) ^{2} - 4 \cdot 6 \cdot 99}{2 \cdot 6}

(- 62)^{2} - 4 \cdot 6 \cdot 99 = 2367

x_{1, 2} = \frac{- ( - 62 ) \pm 2 367}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 62 ) + 2 367}{2 \cdot 6}, x_{2} = \frac{- ( - 62 ) - 2 367}{2 \cdot 6}

x = \frac{- ( - 62 ) + 2 367}{2 \cdot 6} : \frac{31 + 367}{6}

x = \frac{- ( - 62 ) - 2 367}{2 \cdot 6} : \frac{31 - 367}{6}

The solutions to the quadratic equation are:

x = \frac{31 + 367}{6}, x = \frac{31 - 367}{6}

- 2 - 4 w = 7 w - 8

62

0 = 5 x^{2} + x + 1

3 x^{2} + 2 x - 1

- \frac{1}{3} x - 6 = - 3

What are the solutions to the equation 6x^2-62x+99=0 ?
The solutions to the equation 6x^2-62x+99=0 are x=(31+sqrt(367))/6 ,x=(31-sqrt(367))/6
Find the zeros of 6x^2-62x+99=0
The zeros of 6x^2-62x+99=0 are x=(31+sqrt(367))/6 ,x=(31-sqrt(367))/6