What are the solutions to the equation -0.3x^2+0.6x=22 ?

The solutions to the equation -0.3x^2+0.6x=22 are x=1-isqrt(217/3),x=1+isqrt(217/3)

Find the zeros of -0.3x^2+0.6x=22

The zeros of -0.3x^2+0.6x=22 are x=1-isqrt(217/3),x=1+isqrt(217/3)

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- 0.3 x^{2} + 0.6 x = 22

x = 1 - i \frac{217}{3}, x = 1 + i \frac{217}{3}

Solution steps

- 0.3 x^{2} + 0.6 x = 22

Multiply both sides by

10

- 3 x^{2} + 6 x = 220

Move

220

to the left side

- 3 x^{2} + 6 x - 220 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 6 \pm 6 ^{2} - 4 ( - 3 ) ( - 220 )}{2 ( - 3 )}

Simplify

6^{2} - 4 (- 3) (- 220) : 2651 i

x_{1, 2} = \frac{- 6 \pm 2 651 i}{2 ( - 3 )}

Separate the solutions

x_{1} = \frac{- 6 + 2 651 i}{2 ( - 3 )}, x_{2} = \frac{- 6 - 2 651 i}{2 ( - 3 )}

x = \frac{- 6 + 2 651 i}{2 ( - 3 )} : 1 - i \frac{217}{3}

x = \frac{- 6 - 2 651 i}{2 ( - 3 )} : 1 + i \frac{217}{3}

The solutions to the quadratic equation are:

x = 1 - i \frac{217}{3}, x = 1 + i \frac{217}{3}

x + 3 = 9 - x^{2}

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

- 4 b^{2} - 3 b + 14 = 4

x^{2} - 2.4 x = 2.2

t^{2} - 6 = 0

What are the solutions to the equation -0.3x^2+0.6x=22 ?
The solutions to the equation -0.3x^2+0.6x=22 are x=1-isqrt(217/3),x=1+isqrt(217/3)
Find the zeros of -0.3x^2+0.6x=22
The zeros of -0.3x^2+0.6x=22 are x=1-isqrt(217/3),x=1+isqrt(217/3)