2x-5=3x^2 find the root of X

For this case we must solve the following quadratic equation:[tex]3x ^ 2-2x + 5 = 0[/tex]Where:[tex]a = 3\\b = -2\\c = 5[/tex]The roots are given by:[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2a}[/tex]Substituting the values we have:[tex]x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {2 \pm \sqrt {4-60}} {6}\\x = \frac {2 \pm \sqrt {-56}} {6}[/tex]By definition we have to:[tex]i ^ 2 = -1\\x = \frac {2 \pm \sqrt {56i ^ 2}} {6}\\x = \frac {2 \pm i \sqrt {56}} {6}\\x = \frac {2 \pm i \sqrt {2 ^ 2 * 14}} {6}\\x = \frac {2 \pm 2i \sqrt {14}} {6}\\x = \frac {1 \pm i \sqrt {14}} {3}[/tex]We have two complex roots:[tex]x_ {1} = \frac {1+ i \sqrt {14}} {3}\\x_ {2} = \frac {1- i \sqrt {14}} {3}[/tex]Answer:[tex]x_ {1} = \frac {1+ i \sqrt {14}} {3}\\x_ {2} = \frac {1- i \sqrt {14}} {3}[/tex]