## Algebra 1

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