What are the solutions to the equation 10-6x-9x^2=-9x ?

The solutions to the equation 10-6x-9x^2=-9x are x=-(-1+sqrt(41))/6 ,x=(1+sqrt(41))/6

Find the zeros of 10-6x-9x^2=-9x

The zeros of 10-6x-9x^2=-9x are x=-(-1+sqrt(41))/6 ,x=(1+sqrt(41))/6

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10 - 6 x - 9 x^{2} = - 9 x

x = - \frac{- 1 + 41}{6}, x = \frac{1 + 41}{6}

Decimal Notation

x = - 0.90052 \dots, x = 1.23385 \dots

Solution steps

10 - 6 x - 9 x^{2} = - 9 x

Move

9 x

to the left side

- 9 x^{2} + 3 x + 10 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 3 \pm 3 ^{2} - 4 ( - 9 ) \cdot 10}{2 ( - 9 )}

3^{2} - 4 (- 9) \cdot 10 = 341

x_{1, 2} = \frac{- 3 \pm 3 41}{2 ( - 9 )}

Separate the solutions

x_{1} = \frac{- 3 + 3 41}{2 ( - 9 )}, x_{2} = \frac{- 3 - 3 41}{2 ( - 9 )}

x = \frac{- 3 + 3 41}{2 ( - 9 )} : - \frac{- 1 + 41}{6}

x = \frac{- 3 - 3 41}{2 ( - 9 )} : \frac{1 + 41}{6}

The solutions to the quadratic equation are:

x = - \frac{- 1 + 41}{6}, x = \frac{1 + 41}{6}

\frac{m}{m ^{2}}

- - 28

x^{2} z^{2} - 4 z^{2} + x^{4} - 4 x^{2}

x + 10 > 12

(\frac{1}{4})^{7}

What are the solutions to the equation 10-6x-9x^2=-9x ?
The solutions to the equation 10-6x-9x^2=-9x are x=-(-1+sqrt(41))/6 ,x=(1+sqrt(41))/6
Find the zeros of 10-6x-9x^2=-9x
The zeros of 10-6x-9x^2=-9x are x=-(-1+sqrt(41))/6 ,x=(1+sqrt(41))/6