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

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

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

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

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

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

Move

10

to the left side

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}

5 x + 8 = 6 x

- 8 ∣ - x - 6 ∣ + 10 = 2

w^{5} + 2 \geq - 1

8 x^{2} - 6 x - 2

2 x^{3} - 50 x

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