What are the solutions to the equation 3w^2-4w+5=0 ?

The solutions to the equation 3w^2-4w+5=0 are w= 2/3+i(sqrt(11))/3 ,w= 2/3-i(sqrt(11))/3

Find the zeros of 3w^2-4w+5=0

The zeros of 3w^2-4w+5=0 are w= 2/3+i(sqrt(11))/3 ,w= 2/3-i(sqrt(11))/3

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3 w^{2} - 4 w + 5 = 0

w = \frac{2}{3} + i \frac{11}{3}, w = \frac{2}{3} - i \frac{11}{3}

Solution steps

3 w^{2} - 4 w + 5 = 0

Solve with the quadratic formula

w_{1, 2} = \frac{- ( - 4 ) \pm ( - 4 ) ^{2} - 4 \cdot 3 \cdot 5}{2 \cdot 3}

Simplify

(- 4)^{2} - 4 \cdot 3 \cdot 5 : 211 i

w_{1, 2} = \frac{- ( - 4 ) \pm 2 11 i}{2 \cdot 3}

Separate the solutions

w_{1} = \frac{- ( - 4 ) + 2 11 i}{2 \cdot 3}, w_{2} = \frac{- ( - 4 ) - 2 11 i}{2 \cdot 3}

w = \frac{- ( - 4 ) + 2 11 i}{2 \cdot 3} : \frac{2}{3} + i \frac{11}{3}

w = \frac{- ( - 4 ) - 2 11 i}{2 \cdot 3} : \frac{2}{3} - i \frac{11}{3}

The solutions to the quadratic equation are:

w = \frac{2}{3} + i \frac{11}{3}, w = \frac{2}{3} - i \frac{11}{3}

625 x - 100 x^{2} + 4 x^{3} > 1000

2 a^{2} b - 4 a b^{2}

\frac{1}{3} (2 x + 8) + \frac{4 x - 8}{6} = - x + \frac{7}{2}

\frac{5 ^{2} 5 ^{3}}{5 ^{3} 5 ^{2}}

e^{- \frac{( - 1 )}{\frac{100}{10}}}

What are the solutions to the equation 3w^2-4w+5=0 ?
The solutions to the equation 3w^2-4w+5=0 are w= 2/3+i(sqrt(11))/3 ,w= 2/3-i(sqrt(11))/3
Find the zeros of 3w^2-4w+5=0
The zeros of 3w^2-4w+5=0 are w= 2/3+i(sqrt(11))/3 ,w= 2/3-i(sqrt(11))/3