For this case, we have the following quadratic equation:
[tex]16x ^ 2 + 24x + 5 = 5\\16x ^ 2 + 24x + 5-5 = 0\\16x ^ 2 + 24x = 0[/tex]
If we divide between 4 on both sides to simplify we have:
[tex]4x ^ 2 + 6x = 0[/tex]
This equation is of the form:
[tex]ax ^ 2 + bx + c = 0[/tex]
Where:
[tex]a = 4\\b = 6\\c = 0[/tex]
Its roots are given by:
[tex]x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\x = \frac {-6 \pm \sqrt {6 ^ 2-4 (4) (0)}} {2 (4)}\\x = \frac {-6 \pm \sqrt {36}} {8}\\x = \frac {-6 \pm6} {8}[/tex]
So, we have two roots:
[tex]x_ {1} = \frac {-6 + 6} {8} = 0\\x_ {2} = \frac {-6-6} {8} = - \frac {12} {8} = - \frac {3} {2}[/tex]
Answer:
The roots are:[tex]x_ {1} = 0\ and\ x_ {2} = - \frac {3} {2}[/tex]
None of the options given are solution
