What are the solutions to the equation -3x^2+6x=-321 ?

The solutions to the equation -3x^2+6x=-321 are x=1-6sqrt(3),x=1+6sqrt(3)

Find the zeros of -3x^2+6x=-321

The zeros of -3x^2+6x=-321 are x=1-6sqrt(3),x=1+6sqrt(3)

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- 3 x^{2} + 6 x = - 321

x = 1 - 63, x = 1 + 63

Decimal Notation

x = - 9.39230 \dots, x = 11.39230 \dots

Solution steps

- 3 x^{2} + 6 x = - 321

Move

321

to the left side

- 3 x^{2} + 6 x + 321 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 3 ) \cdot 321}{2 ( - 3 )}

6^{2} - 4 (- 3) \cdot 321 = 363

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

Separate the solutions

x_{1} = \frac{- 6 + 36 3}{2 ( - 3 )}, x_{2} = \frac{- 6 - 36 3}{2 ( - 3 )}

x = \frac{- 6 + 36 3}{2 ( - 3 )} : 1 - 63

x = \frac{- 6 - 36 3}{2 ( - 3 )} : 1 + 63

The solutions to the quadratic equation are:

x = 1 - 63, x = 1 + 63

300 = 18 x^{2} - 180 x - 3300

- 0.05 x^{2} + 16 x + 80 = 0

x, z = - x^{2} - y^{2}

x^{2} - (3 + 1) x + 3 = 0

- 6 x^{2} + 9 x + 24 = 0

What are the solutions to the equation -3x^2+6x=-321 ?
The solutions to the equation -3x^2+6x=-321 are x=1-6sqrt(3),x=1+6sqrt(3)
Find the zeros of -3x^2+6x=-321
The zeros of -3x^2+6x=-321 are x=1-6sqrt(3),x=1+6sqrt(3)