Understanding the Quadratic Equation: Simplifying y = 3(2)² – 12(2) + 15 Step by Step

Solving quadratic equations is a foundational skill in algebra, and one that often causes confusion—even for experienced learners. In this article, we break down the equation:

y = 3(2)² – 12(2) + 15
and simplify it step by step to reveal why it finally simplifies to y = 3. Whether you're studying algebra, teacher, or self-learner, understanding how to evaluate and simplify expressions—especially quadratic forms—is essential for mastering more complex math concepts.

Understanding the Context

What Is the Given Equation?

Our starting equation is:
y = 3(2)² – 12(2) + 15

At first glance, this seems like a standard quadratic expression—but in this case, we’ll treat it like a linear evaluation problem since no variable “y” is isolated as the subject. But the challenge lies not in solving for a variable, but in correctly evaluating the expression using order of operations (PEMDAS/BODMAS).

1. $12y^3 - y^5 3y^2$ 2. $ {6x^5 - | StudyX

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1. $12y^3 - y^5 3y^2$ 2. $ {6x^5 - | StudyX
Solved: (5 y+1)^2-(3 y+4)(2-3 y) [Math]
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Solution to problem # 2.3.15
Solved: y^3-4 y^2+3 y+1 from 6 y^3-4 y^2+5 y+5 [Math]

Key Insights

Step 1: Apply the Exponent First

Because of the exponent (2) in (2)², we apply the P (Parentheses) rule in sequence with E (Exponents):
(2)² = 4

Now substitute:
y = 3 × 4 – 12 × 2 + 15

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Final Thoughts

Step 2: Perform Multiplication from Left to Right

Multiplication comes before addition and subtraction, per BODMAS/PEMDAS:

  • First term: 3 × 4 = 12
  • Second term: 12 × 2 = 24

Now substitute:
y = 12 – 24 + 15

Step 3: Evaluate from Left to Right (Addition and Subtraction)

We proceed left to right:

  • 12 – 24 = –12
  • Then: –12 + 15 = 3

Final Result:

y = 3