What are the solutions to the equation 6n^2-14=11n ?

The solutions to the equation 6n^2-14=11n are n=(11+sqrt(457))/(12),n=(11-sqrt(457))/(12)

Find the zeros of 6n^2-14=11n

The zeros of 6n^2-14=11n are n=(11+sqrt(457))/(12),n=(11-sqrt(457))/(12)

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6 n^{2} - 14 = 11 n

n = \frac{11 + 457}{12}, n = \frac{11 - 457}{12}

Solution steps

6 n^{2} - 14 = 11 n

Move

11 n

to the left side

6 n^{2} - 14 - 11 n = 0

Write in the standard form

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

6 n^{2} - 11 n - 14 = 0

Solve with the quadratic formula

n_{1, 2} = \frac{- ( - 11 ) \pm ( - 11 ) ^{2} - 4 \cdot 6 ( - 14 )}{2 \cdot 6}

(- 11)^{2} - 4 \cdot 6 (- 14) = 457

n_{1, 2} = \frac{- ( - 11 ) \pm 457}{2 \cdot 6}

Separate the solutions

n_{1} = \frac{- ( - 11 ) + 457}{2 \cdot 6}, n_{2} = \frac{- ( - 11 ) - 457}{2 \cdot 6}

n = \frac{- ( - 11 ) + 457}{2 \cdot 6} : \frac{11 + 457}{12}

n = \frac{- ( - 11 ) - 457}{2 \cdot 6} : \frac{11 - 457}{12}

The solutions to the quadratic equation are:

n = \frac{11 + 457}{12}, n = \frac{11 - 457}{12}

3 (a + b)^{2} + 11 (a + b) - 4

36 x \leq x^{3}

\frac{1}{3 ^{- 5}}

2 + z = 6 z^{2}

- 17 = \frac{x}{5}

What are the solutions to the equation 6n^2-14=11n ?
The solutions to the equation 6n^2-14=11n are n=(11+sqrt(457))/(12),n=(11-sqrt(457))/(12)
Find the zeros of 6n^2-14=11n
The zeros of 6n^2-14=11n are n=(11+sqrt(457))/(12),n=(11-sqrt(457))/(12)