## Algebra 1

$x = 6$ and $x = -1$
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. $x^2 - 5x - 6 = 0$ Step 2: Determine our a, b, and c values $a = 1, b = -5, c = -6$ Step 3: Evaluate $b^2 - 4ac$, using our values of a, b, and c $b^2 - 4ac$ = $(-5)^2 - 4(1)(-6) = 25 + 24 = 49$ Based on our rules of the discriminant, since $49>0$, 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. $x^2 - 5x - 6 = 0$ Step 2: Determine our a, b, and c values $a = 1, b = -5, c = -6$ Step 3: Substitute our values of a, b, and c into the formula $x = \frac{-b \frac{+}{-} \sqrt(b^2 - 4ac)}{2a}$ $x = \frac{-(-5) \frac{+}{-} \sqrt((-5)^2 - 4(1)(-6)}{2(1)}$ $x = \frac{5 \frac{+}{-} \sqrt(25 + 24)}{2}$ $x = \frac{5 \frac{+}{-} \sqrt(49)}{2}$ $x = \frac{5 \frac{+}{-} 7}{2}$ $x = \frac{12}{2}$ and $x = \frac{-2}{2}$ therefore, $x = 6$ and $x = -1$