Question

The roots of the equation 3x^2-4x+2=0​

Answer 1

For this case we must find the roots of the following equation:

$3x ^ 2-4x + 2 = 0$

We have that the roots will come from:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}$

Where:

$a = 3\\b = -4\\c = 2$

Substituting the values:

$x = \frac {- (- 4) \pm \sqrt {(- 4) ^ 2-4 (3) (2)}} {2 (3)}\\x = \frac {4 \pm \sqrt {16-24}} {6}\\x = \frac {4 \pm \sqrt {-8}} {6}\\x = \frac {4 \pm \sqrt {-1 * 8}} {6}\\x = \frac {4 \pmi \sqrt {2 ^ 2 * 2}} {6}\\x = \frac {4 \pm2i \sqrt {2}} {6}\\x = \frac {2 \pmi \sqrt {2}} {3}$

We have two complex roots:

$x_ {1} = \frac {2 + i \sqrt {2}} {3}\\x_ {2} = \frac {2-i \sqrt {2}} {3}$

Answer:

$x_ {1} = \frac {2 + i \sqrt {2}} {3}\\x_ {2} = \frac {2-i \sqrt {2}} {3}$