1. **Problem Statement:** Find the values of $x$ such that the area of the triangle with vertices $(x,4)$, $(2,-6)$, and $(5,4)$ is 35 square units. 2. **Formula for Area of Triangle with Coordinates:** The area $A$ of a triangle with vertices $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ is given by: $$ A = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| $$ 3. **Substitute the given points:** Let $(x_1,y_1) = (x,4)$, $(x_2,y_2) = (2,-6)$, $(x_3,y_3) = (5,4)$. Calculate the expression inside the absolute value: $$ x( -6 - 4 ) + 2(4 - 4) + 5(4 - (-6)) = x(-10) + 2(0) + 5(10) = -10x + 50 $$ 4. **Set up the area equation:** $$ 35 = \frac{1}{2} | -10x + 50 | $$ Multiply both sides by 2: $$ 70 = | -10x + 50 | $$ 5. **Solve the absolute value equation:** $$ -10x + 50 = 70 \quad \text{or} \quad -10x + 50 = -70 $$ For the first case: $$ -10x + 50 = 70 \\ -10x = 20 \\ x = \frac{20}{-10} = -2 $$ For the second case: $$ -10x + 50 = -70 \\ -10x = -120 \\ x = \frac{-120}{-10} = 12 $$ 6. **Final answer:** The values of $x$ are $12$ and $-2$. **Answer choice:** (a) 12, −2