Solve the equation after making an appropriate substitution. Complete parts a through d. 

left parenthesis 13 x minus 3 right parenthesis squared minus 2 left parenthesis 13 x minus 3 right parenthesis minus 15 equals 0(13x−3)2−2(13x−3)−15=0

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Part 1

​a) Determine the appropriate substitution using the new variable u. 

uequals=13 x minus 313x−3

Part 2

​b) Write the new quadratic equation using the variable. 

u squared minus 2 u minus 15u2−2u−15equals=0

​(Simplify your answer. Do not​ factor.)

Part 3

​c) Solve the new equation for the variable u. 

uequals=5 comma negative 35,−3

​(Simplify your answer. Type your​ answer(s) as integers or simplified fractions. Use a comma to separate answers as​ needed.)

Part 4

​d) Determine the solution to the original equation.

The solution set to the original equation is 

StartSet nothing EndSetenter your response here.

​(Simplify your answer. Type your​ answer(s) as integers or simplified fractions. Use a comma to separate answers as​ needed.)

Let's go step-by-step and work through each part of the problem.

### Part 1: Substitution
You are asked to make a substitution.

Given equation:  
\[
(13x - 3)^2 - 2(13x - 3) - 15 = 0
\]
We are asked to use the substitution:
\[
u = 13x - 3
\]
This is our appropriate substitution for the variable \( u \).

### Part 2: Rewrite the equation
Now, rewrite the given equation in terms of \( u \).

Substitute \( u \) for \( 13x - 3 \):
\[
u^2 - 2u - 15 = 0
\]
This is the quadratic equation using the new variable \( u \).

### Part 3: Solve for \( u \)
We now solve the quadratic equation:
\[
u^2 - 2u - 15 = 0
\]
To solve, use the quadratic formula:
\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -2 \), and \( c = -15 \).

Calculate the discriminant:
\[
b^2 - 4ac = (-2)^2 - 4(1)(-15) = 4 + 60 = 64
\]
Now apply the quadratic formula:
\[
u = \frac{-(-2) \pm \sqrt{64}}{2(1)} = \frac{2 \pm 8}{2}
\]
This gives two solutions:
\[
u = \frac{2 + 8}{2} = \frac{10}{2} = 5
\]
and
\[
u = \frac{2 - 8}{2} = \frac{-6}{2} = -3
\]
So, \( u = 5 \) and \( u = -3 \).

### Part 4: Solve for \( x \)
Now, we substitute back for \( u \).

Recall that \( u = 13x - 3 \), so we now solve the two equations:
1. \( 13x - 3 = 5 \)
2. \( 13x - 3 = -3 \)

For the first equation:
\[
13x - 3 = 5
\]
Add 3 to both sides:
\[
13x = 8
\]
Now divide by 13:
\[
x = \frac{8}{13}
\]

For the second equation:
\[
13x - 3 = -3
\]
Add 3 to both sides:
\[
13x = 0
\]
Now divide by 13:
\[
x = 0
\]

### Final Answer:
The solution set to the original equation is:
\[
\left\{ 0, \frac{8}{13} \right\}
\]

Would you like more details or have any questions?  
Here are 5 related questions you might consider:
1. How do you apply the quadratic formula for different types of quadratic equations?
2. Can you solve a quadratic equation using factoring instead of the quadratic formula?
3. How do you check if your solutions for \( x \) are correct by substituting them back into the original equation?
4. What happens if the discriminant is negative, and how do you solve such equations?
5. How does substitution simplify solving complex algebraic equations?

**Tip:** Always check your solutions by substituting them back into the original equation to ensure they satisfy it.

Solve the equation after making an appropriate substitution. Complete parts a through d. (13x−3)^2−2(13x−3)−15=0.

Part 1: Determine the appropriate substitution using the new variable u. u = 13x−3

Part 2: Write the new quadratic equation using the variable. u^2−2u−15=0

Part 3: Solve the new equation for the variable u. u = 5, -3

Part 4: Determine the solution to the original equation.

Step-by-step guide to solving the quadratic equation (13x−3)^2−2(13x−3)−15=0 using substitution. Learn how to transform and solve quadratic equations using algebraic methods and the quadratic formula. Suitable for students in Grades 9-10.

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