## Algebra 1

x = -3 + $2\sqrt 2$ x = -3 - $2\sqrt 2$
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. $x^2 + 6x + 1 = 0$ Step 2: Determine our a, b, and c values $a = 1, b = 6, c = 1$ Step 3: Evaluate $b^2 - 4ac$, using our values of a, b, and c $b^2 - 4ac$ = $6^2 - 4(1)(1) = 36 - 4 = 32$ Based on our rules of the discriminant, since $32>0$, there are two real number solutions. Now, using the Quadratic Formula, we can find these two real number solutions. Step 1 and 2 are already complete from finding the discriminant. Step 1: Write the equation in standard form. $x^2 + 6x + 1 = 0$ Step 2: Determine our a, b, and c values $a = 1, b = 6, c = 1$ Step 3: Substitute our values of a, b, and c into the formula $x = \frac{-b \frac{+}{-} \sqrt(b^2 - 4ac)}{2a}$ $x = \frac{-6 \frac{+}{-} \sqrt(6^2 - 4(1)(1)}{2(1)}$ $x = \frac{-6 \frac{+}{-} \sqrt(36-4)}{2}$ $x = \frac{-6 \frac{+}{-} \sqrt(32)}{2}$ $x = \frac{-6 \frac{+}{-} 4 \sqrt2}{2}$ $x = -3 \frac{+}{-} 2 \sqrt2$ therefore, $x = -3 + 2 \sqrt2$ and $x = -3 - 2 \sqrt2$