#### Answer

$x = -\frac{1}{3}$ and $x = -\frac{1}{2}$

#### Work Step by Step

First, we should look at the discriminant ($b^2 - 4ac$) to determine how many real number solutions we have. The rules are such:
$(b^2 - 4ac) \gt 0$ then there are two real number solutions
$(b^2 - 4ac) \lt 0$ then there is one real number solution
$(b^2 - 4ac) = 0$ then there is no real number solution
Evaluating the discriminant:
Step 1: Write the equation in standard form.
$6x^2+5x+1 = 0$
Step 2: Determine our a, b, and c values
$a = 6, b = 5, c = 1$
Step 3: Evaluate $b^2 - 4ac$, using our values of a, b, and c
$b^2 - 4ac$ = $5^2 - 4(6)(1) = 25 - 24 = 1$
Based on our rules of the discriminant, since $1 > 0$, there are two real number solutions.
Now, using the Quadratic Formula, we can find this real number solution. Step 1 and 2 are already complete from finding the discriminant.
Step 1: Write the equation in standard form.
$6x^2 + 5x +1 = 0$
Step 2: Determine our a, b, and c values
$a = 6, b = 5, c = 1$
Step 3: Substitute our values of a, b, and c into the formula
$x = \frac{-b \frac{+}{-} \sqrt(b^2 - 4ac)}{2a}$
$x = \frac{-5 \frac{+}{-} \sqrt(5^2 - 4(6)(1)}{2(6)}$
$x = \frac{-5 \frac{+}{-} \sqrt(25 - 24)}{12}$
$x = \frac{-5 \frac{+}{-} \sqrt(1)}{12}$
$x = \frac{-5 \frac{+}{-} 1}{12}$
$x = \frac{-4}{12}$ and $x = \frac{-6}{12}$
therefore,
$x = -\frac{1}{3}$ and $x = -\frac{1}{2}$
We have found our real number solutions.