What are the solutions to the equation (3q^2-12q+20)=0 ?

The solutions to the equation (3q^2-12q+20)=0 are q=2+i(2sqrt(6))/3 ,q=2-i(2sqrt(6))/3

Find the zeros of (3q^2-12q+20)=0

The zeros of (3q^2-12q+20)=0 are q=2+i(2sqrt(6))/3 ,q=2-i(2sqrt(6))/3

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(3 q^{2} - 12 q + 20) = 0

q = 2 + i \frac{2 6}{3}, q = 2 - i \frac{2 6}{3}

Solution steps

(3 q^{2} - 12 q + 20) = 0

Refine

3 q^{2} - 12 q + 20 = 0

Solve with the quadratic formula

q_{1, 2} = \frac{- ( - 12 ) \pm ( - 12 ) ^{2} - 4 \cdot 3 \cdot 20}{2 \cdot 3}

Simplify

(- 12)^{2} - 4 \cdot 3 \cdot 20 : 46 i

q_{1, 2} = \frac{- ( - 12 ) \pm 4 6 i}{2 \cdot 3}

Separate the solutions

q_{1} = \frac{- ( - 12 ) + 4 6 i}{2 \cdot 3}, q_{2} = \frac{- ( - 12 ) - 4 6 i}{2 \cdot 3}

q = \frac{- ( - 12 ) + 4 6 i}{2 \cdot 3} : 2 + i \frac{2 6}{3}

q = \frac{- ( - 12 ) - 4 6 i}{2 \cdot 3} : 2 - i \frac{2 6}{3}

The solutions to the quadratic equation are:

q = 2 + i \frac{2 6}{3}, q = 2 - i \frac{2 6}{3}

\frac{4 x}{2} < 8 x + 3

3 k^{2} - 10 k + 7

- y^{2} + 2 y + 3 = 0

p^{2} + 20 p + 19

\frac{7}{10} \cdot \frac{1}{2}

What are the solutions to the equation (3q^2-12q+20)=0 ?
The solutions to the equation (3q^2-12q+20)=0 are q=2+i(2sqrt(6))/3 ,q=2-i(2sqrt(6))/3
Find the zeros of (3q^2-12q+20)=0
The zeros of (3q^2-12q+20)=0 are q=2+i(2sqrt(6))/3 ,q=2-i(2sqrt(6))/3