Question

What are the solutions of the equation?

Answer 1

For this case, we have the following quadratic equation:

$16x ^ 2 + 24x + 5 = 5\\16x ^ 2 + 24x + 5-5 = 0\\16x ^ 2 + 24x = 0$

If we divide between 4 on both sides to simplify we have:

$4x ^ 2 + 6x = 0$

 This equation is of the form:

$ax ^ 2 + bx + c = 0$

Where:

$a = 4\\b = 6\\c = 0$

Its roots are given by:

$x = \frac {-b \pm \sqrt {b ^ 2-4 (a) (c)}} {2 (a)}\\x = \frac {-6 \pm \sqrt {6 ^ 2-4 (4) (0)}} {2 (4)}\\x = \frac {-6 \pm \sqrt {36}} {8}\\x = \frac {-6 \pm6} {8}$

So, we have two roots:

$x_ {1} = \frac {-6 + 6} {8} = 0\\x_ {2} = \frac {-6-6} {8} = - \frac {12} {8} = - \frac {3} {2}$

Answer:

The roots are:$x_ {1} = 0\ and\ x_ {2} = - \frac {3} {2}$

None of the options given are solution

Answer 2

Answer:

B

Step-by-step explanation:

you have the wrong equation posted.

if you meant to post $16x^2+24x+5=0$

the answer would be B.