What are the solutions to the equation 3x-9x^2+10=0 ?

The solutions to the equation 3x-9x^2+10=0 are x=-(-1+sqrt(41))/6 ,x=(1+sqrt(41))/6

Find the zeros of 3x-9x^2+10=0

The zeros of 3x-9x^2+10=0 are x=-(-1+sqrt(41))/6 ,x=(1+sqrt(41))/6

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

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

Decimal Notation

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

Solution steps

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

Write in the standard form

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

- 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}

x + 8 \leq 13

0.534 = 4.9 t^{2} + 0.442 t

(x - 3)^{2} - (x + 4)^{2}

\frac{3 p + 1}{2} - \frac{2 p}{5} = 6

\frac{9}{2 - i}

What are the solutions to the equation 3x-9x^2+10=0 ?
The solutions to the equation 3x-9x^2+10=0 are x=-(-1+sqrt(41))/6 ,x=(1+sqrt(41))/6
Find the zeros of 3x-9x^2+10=0
The zeros of 3x-9x^2+10=0 are x=-(-1+sqrt(41))/6 ,x=(1+sqrt(41))/6