For this case we have the following quadratic equation:
[tex]3x ^ 2-7x + 3 = 0[/tex]
The solutions will be given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]
Where:
[tex]a = 3\\b = -7\\c = 3[/tex]
Substituting the values we have:
[tex]x = \frac {- (- 7) \pm \sqrt {(- 7) ^ 2-4 (3) (3)}} {2 (3)}\\x = \frac {7 \pm \sqrt {49-36}} {6}\\x = \frac {7 \pm \sqrt {13}} {6}[/tex]
We have two roots:
[tex]x_ {1} = \frac {7- \sqrt {13}} {6}\\x_ {2} = \frac {7+ \sqrt{13}} {6}[/tex]
ANswer:
[tex]x_ {1} = \frac {7- \sqrt {13}} {6}\\x_ {2} = \frac {7+ \sqrt{13}} {6}[/tex]
