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------------------------------------------------ Solve the equation using the quadratic formula:

[tex]15x^2 + 13x = 0[/tex]

a. [tex]x = -\frac{13}{15}, 0[/tex]
b. [tex]x = 0[/tex]
c. [tex]x = \frac{13}{15}, 0[/tex]
d. [tex]x = \pm \frac{13}{15}[/tex]

Please select the best answer from the choices provided:
A
B
C
D

To solve the quadratic equation [tex]$15x^2 + 13x = 0$[/tex], we can use the quadratic formula. The general form of a quadratic equation is [tex]$ax^2 + bx + c = 0$[/tex]. For this equation:

- [tex]$a = 15$[/tex]
- [tex]$b = 13$[/tex]
- [tex]$c = 0$[/tex]

The quadratic formula is given by:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

However, since [tex]$c = 0$[/tex], the equation [tex]$15x^2 + 13x = 0$[/tex] can be factored easily. We can factor out an [tex]$x$[/tex]:

[tex]\[ x(15x + 13) = 0 \][/tex]

This gives us two solutions:

1. [tex]$ x = 0 $[/tex]

2. [tex]$ 15x + 13 = 0 $[/tex]

To solve [tex]$15x + 13 = 0$[/tex], subtract 13 from both sides:

[tex]\[ 15x = -13 \][/tex]

Divide both sides by 15:

[tex]\[ x = -\frac{13}{15} \][/tex]

So, the solutions are [tex]$x = 0$[/tex] and [tex]$x = -\frac{13}{15}$[/tex].

Among the options provided:

- a. [tex]$x = -\frac{13}{15}, 0$[/tex]
- b. [tex]$x = 0$[/tex]
- c. [tex]$x = \frac{13}{15}, 0$[/tex]
- d. [tex]$x = \pm \frac{13}{15}$[/tex]

The correct choice for the solutions is option A: [tex]$x = -\frac{13}{15}, 0$[/tex].