What are the solutions to the equation -3x^2+6x=-1 ?

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

Find the zeros of -3x^2+6x=-1

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

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- 3 x^{2} + 6 x = - 1

x = - \frac{- 3 + 2 3}{3}, x = \frac{3 + 2 3}{3}

Solution steps

- 3 x^{2} + 6 x = - 1

Move

1

to the left side

- 3 x^{2} + 6 x + 1 = 0

Solve with the quadratic formula

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

6^{2} - 4 (- 3) \cdot 1 = 43

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

Separate the solutions

x_{1} = \frac{- 6 + 4 3}{2 ( - 3 )}, x_{2} = \frac{- 6 - 4 3}{2 ( - 3 )}

x = \frac{- 6 + 4 3}{2 ( - 3 )} : - \frac{- 3 + 2 3}{3}

x = \frac{- 6 - 4 3}{2 ( - 3 )} : \frac{3 + 2 3}{3}

The solutions to the quadratic equation are:

x = - \frac{- 3 + 2 3}{3}, x = \frac{3 + 2 3}{3}

(- 2)^{3} = (- 2) x (- 2) x (- 2)

4 s^{2} + 4 s + 5 = 0

0 = - 2.7 t^{2} + 30 t + 6.5

f^{2} + 7 f + 10 = 0

(- 5)^{2} + (2)^{2} = D^{2}

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