## Algebra 1

$x = 8$ and $x = -3$
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. $(2x-5)(2x-5) = 121$ $4x^4 -10x^2 - 10x^2 +25 - 121 = 0$ Use FOIL to simplify $4x^2 - 20x^2 - 96 = 0$ Combine like terms Step 2: Determine our a, b, and c values $a = 4, b = -20, c = -96$ Step 3: Evaluate $b^2 - 4ac$, using our values of a, b, and c $b^2 - 4ac$ = $(-20)^2 - 4(4)(-96) =400 + 1536 = 1936$ Based on our rules of the discriminant, since $1936 > 0$, there are two real number solutions. Now, using the Quadratic Formula, we can find these real number solutions. Step 1 and 2 are already complete from finding the discriminant. Step 1: Write the equation in standard form. $(2x-5)(2x-5) = 121$ $4x^4 -10x^2 - 10x^2 +25 - 121 = 0$ Use FOIL to simplify $4x^2 - 20x^2 - 96 = 0$ Combine like terms Step 2: Determine our a, b, and c values $a = 4, b = -20, c = -96$ Step 3: Substitute our values of a, b, and c into the formula $x = \frac{-b \frac{+}{-} \sqrt(b^2 - 4ac)}{2a}$ $x = \frac{-(-20) \frac{+}{-} \sqrt((-20)^2 - 4(4)(-96)}{2(4)}$ $x = \frac{20 \frac{+}{-} \sqrt(400 + 1536)}{8}$ $x = \frac{20 \frac{+}{-} \sqrt(1936)}{8}$ $x = \frac{20 \frac{+}{-} 44}{8}$ $x = \frac{64}{8}$ and $x = \frac{-24}{8}$ therefore, $x = 8$ and $x = -3$ We have found our real number solutions.