What are the solutions to the equation 6x^2-14x-98=0 ?

The solutions to the equation 6x^2-14x-98=0 are x=(7(1+sqrt(13)))/6 ,x=(7(1-sqrt(13)))/6

Find the zeros of 6x^2-14x-98=0

The zeros of 6x^2-14x-98=0 are x=(7(1+sqrt(13)))/6 ,x=(7(1-sqrt(13)))/6

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6 x^{2} - 14 x - 98 = 0

x = \frac{7 ( 1 + 13 )}{6}, x = \frac{7 ( 1 - 13 )}{6}

Decimal Notation

x = 5.37314 \dots, x = - 3.03980 \dots

Solution steps

6 x^{2} - 14 x - 98 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- ( - 14 ) \pm ( - 14 ) ^{2} - 4 \cdot 6 ( - 98 )}{2 \cdot 6}

(- 14)^{2} - 4 \cdot 6 (- 98) = 1413

x_{1, 2} = \frac{- ( - 14 ) \pm 14 13}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- ( - 14 ) + 14 13}{2 \cdot 6}, x_{2} = \frac{- ( - 14 ) - 14 13}{2 \cdot 6}

x = \frac{- ( - 14 ) + 14 13}{2 \cdot 6} : \frac{7 ( 1 + 13 )}{6}

x = \frac{- ( - 14 ) - 14 13}{2 \cdot 6} : \frac{7 ( 1 - 13 )}{6}

The solutions to the quadratic equation are:

x = \frac{7 ( 1 + 13 )}{6}, x = \frac{7 ( 1 - 13 )}{6}

\frac{\frac{7}{4} - \frac{2}{3} ( \frac{7}{2} + 1 )}{24 + 2 ( 3 + 2 )}

\frac{( x - 5 ) ^{2}}{x ^{2} - 16} \geq 0

(x - 7)^{2} = 25

3 (x - 4) + 19 = 2 (x + 5)

5 x - 4 \geq 11

What are the solutions to the equation 6x^2-14x-98=0 ?
The solutions to the equation 6x^2-14x-98=0 are x=(7(1+sqrt(13)))/6 ,x=(7(1-sqrt(13)))/6
Find the zeros of 6x^2-14x-98=0
The zeros of 6x^2-14x-98=0 are x=(7(1+sqrt(13)))/6 ,x=(7(1-sqrt(13)))/6