What are the solutions to the equation 6x52x-1=9x-3x15 ?

The solutions to the equation 6x52x-1=9x-3x15 are x=(-9+sqrt(159))/(156),x=-(9+sqrt(159))/(156)

Find the zeros of 6x52x-1=9x-3x15

The zeros of 6x52x-1=9x-3x15 are x=(-9+sqrt(159))/(156),x=-(9+sqrt(159))/(156)

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6 x 52 x - 1 = 9 x - 3 x 15

x = \frac{- 9 + 159}{156}, x = - \frac{9 + 159}{156}

Decimal Notation

x = 0.02313 \dots, x = - 0.13852 \dots

Solution steps

6 x \cdot 52 x - 1 = 9 x - 3 x \cdot 15

Expand

6 x \cdot 52 x - 1 : 312 x^{2} - 1

Expand

9 x - 3 x \cdot 15 : - 36 x

312 x^{2} - 1 = - 36 x

Move

36 x

to the left side

312 x^{2} - 1 + 36 x = 0

Write in the standard form

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

312 x^{2} + 36 x - 1 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 36 \pm 3 6 ^{2} - 4 \cdot 312 ( - 1 )}{2 \cdot 312}

3 6^{2} - 4 \cdot 312 (- 1) = 4159

x_{1, 2} = \frac{- 36 \pm 4 159}{2 \cdot 312}

Separate the solutions

x_{1} = \frac{- 36 + 4 159}{2 \cdot 312}, x_{2} = \frac{- 36 - 4 159}{2 \cdot 312}

x = \frac{- 36 + 4 159}{2 \cdot 312} : \frac{- 9 + 159}{156}

x = \frac{- 36 - 4 159}{2 \cdot 312} : - \frac{9 + 159}{156}

The solutions to the quadratic equation are:

x = \frac{- 9 + 159}{156}, x = - \frac{9 + 159}{156}

y = y^{2} - 56

x^{2} - 3 x - 22 = 0

s^{2} + 40 s + 80 = 0

5 x^{2} - 11 x - 168 = 0

- 16 t^{2} + 32 t + 2 = 0

What are the solutions to the equation 6x52x-1=9x-3x15 ?
The solutions to the equation 6x52x-1=9x-3x15 are x=(-9+sqrt(159))/(156),x=-(9+sqrt(159))/(156)
Find the zeros of 6x52x-1=9x-3x15
The zeros of 6x52x-1=9x-3x15 are x=(-9+sqrt(159))/(156),x=-(9+sqrt(159))/(156)