For this case we have that a quadratic equation is of the form:
[tex]ax ^ 2 + bx + c = 0[/tex]
The roots are given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}[/tex]
We have the following equation:
[tex]4x ^ 2-3x + 9 = 2x + 1\\4x ^ 2-3x-2x + 9-1 = 0\\4x ^ 2-5x + 8 = 0[/tex]
We look for the roots:
[tex]x = \frac {- (- 5) \pm \sqrt {(- 5) ^ 2-4 (4) (8)}} {2 (4)}\\x = \frac {5 \pm \sqrt {25-128}} {8}\\x = \frac {5 \pm \sqrt {-103}} {8}[/tex]
We have to:
[tex]i ^ 2 = -1[/tex]
So:
[tex]x = \frac {5 \pm \sqrt {103i ^ 2}} {8}\\x = \frac {5 \pm i \sqrt {103}} {8}[/tex]
We have two imaginary roots:
[tex]x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}[/tex]
Answer:
[tex]x_ {1} = \frac {5 \ + i \sqrt {103}} {8}\\x_ {2} = \frac {5 \ -i \sqrt {103}} {8}[/tex]
