What are the solutions to the equation 6x^2+3x=1296 ?

The solutions to the equation 6x^2+3x=1296 are x=(-1+sqrt(3457))/4 ,x=-(1+sqrt(3457))/4

Find the zeros of 6x^2+3x=1296

The zeros of 6x^2+3x=1296 are x=(-1+sqrt(3457))/4 ,x=-(1+sqrt(3457))/4

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6 x^{2} + 3 x = 1296

x = \frac{- 1 + 3457}{4}, x = - \frac{1 + 3457}{4}

Decimal Notation

x = 14.44906 \dots, x = - 14.94906 \dots

Solution steps

6 x^{2} + 3 x = 1296

Move

1296

to the left side

6 x^{2} + 3 x - 1296 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 \cdot 6 ( - 1296 )}{2 \cdot 6}

3^{2} - 4 \cdot 6 (- 1296) = 33457

x_{1, 2} = \frac{- 3 \pm 3 3457}{2 \cdot 6}

Separate the solutions

x_{1} = \frac{- 3 + 3 3457}{2 \cdot 6}, x_{2} = \frac{- 3 - 3 3457}{2 \cdot 6}

x = \frac{- 3 + 3 3457}{2 \cdot 6} : \frac{- 1 + 3457}{4}

x = \frac{- 3 - 3 3457}{2 \cdot 6} : - \frac{1 + 3457}{4}

The solutions to the quadratic equation are:

x = \frac{- 1 + 3457}{4}, x = - \frac{1 + 3457}{4}

(s^{2} - 4) = 0

(x - 1) (x + 4) + 4 = 0

\frac{49 + 25 - 9}{2 \cdot 7 \cdot 5} = \frac{49 + 49 - x ^{2}}{2 \cdot 7 \cdot 7}

x^{2} - 6 x - 4 = 0

e^{2} = 5

What are the solutions to the equation 6x^2+3x=1296 ?
The solutions to the equation 6x^2+3x=1296 are x=(-1+sqrt(3457))/4 ,x=-(1+sqrt(3457))/4
Find the zeros of 6x^2+3x=1296
The zeros of 6x^2+3x=1296 are x=(-1+sqrt(3457))/4 ,x=-(1+sqrt(3457))/4