Question

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

Answer 1

For this case we must solve the following quadratic equation:

$3x ^ 2-2x + 5 = 0$

Where:

$a = 3\\b = -2\\c = 5$

The roots are given by:

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

Substituting the values we have:

$x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {- (- 2) \pm \sqrt {(- 2) ^ 2-4 (3) (5)}} {2 (3)}\\x = \frac {2 \pm \sqrt {4-60}} {6}\\x = \frac {2 \pm \sqrt {-56}} {6}$

By definition we have to:

$i ^ 2 = -1\\x = \frac {2 \pm \sqrt {56i ^ 2}} {6}\\x = \frac {2 \pm i \sqrt {56}} {6}\\x = \frac {2 \pm i \sqrt {2 ^ 2 * 14}} {6}\\x = \frac {2 \pm 2i \sqrt {14}} {6}\\x = \frac {1 \pm i \sqrt {14}} {3}$

We have two complex roots:

$x_ {1} = \frac {1+ i \sqrt {14}} {3}\\x_ {2} = \frac {1- i \sqrt {14}} {3}$

Answer:

$x_ {1} = \frac {1+ i \sqrt {14}} {3}\\x_ {2} = \frac {1- i \sqrt {14}} {3}$