What are the solutions to the equation -3w^2+12w-27=0 ?

The solutions to the equation -3w^2+12w-27=0 are w=2-sqrt(5)i,w=2+sqrt(5)i

Find the zeros of -3w^2+12w-27=0

The zeros of -3w^2+12w-27=0 are w=2-sqrt(5)i,w=2+sqrt(5)i

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- 3 w^{2} + 12 w - 27 = 0

w = 2 - 5 i, w = 2 + 5 i

Solution steps

- 3 w^{2} + 12 w - 27 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- 12 \pm 1 2 ^{2} - 4 ( - 3 ) ( - 27 )}{2 ( - 3 )}

Simplify

1 2^{2} - 4 (- 3) (- 27) : 65 i

w_{1, 2} = \frac{- 12 \pm 6 5 i}{2 ( - 3 )}

Separate the solutions

w_{1} = \frac{- 12 + 6 5 i}{2 ( - 3 )}, w_{2} = \frac{- 12 - 6 5 i}{2 ( - 3 )}

w = \frac{- 12 + 6 5 i}{2 ( - 3 )} : 2 - 5 i

w = \frac{- 12 - 6 5 i}{2 ( - 3 )} : 2 + 5 i

The solutions to the quadratic equation are:

w = 2 - 5 i, w = 2 + 5 i

2 x^{2} - 6 x = 7

0 = x^{2} - 13 x + 12

- 0.05 x^{2} + 1.5 x + 1.55 = 0

5 x^{2} + 14 x + 3 = 0

s = 4 π r^{2}, r

What are the solutions to the equation -3w^2+12w-27=0 ?
The solutions to the equation -3w^2+12w-27=0 are w=2-sqrt(5)i,w=2+sqrt(5)i
Find the zeros of -3w^2+12w-27=0
The zeros of -3w^2+12w-27=0 are w=2-sqrt(5)i,w=2+sqrt(5)i