400y_1^2+2×7.54×(1017.87+1017.87) y_1-2×7.54×(1017.87×430+1017.87×70)=0

The given equation is:

\[
400y_1^2 + 2 \times 7.54 \times (1017.87 + 1017.87)y_1 - 2 \times 7.54 \times (1017.87 \times 430 + 1017.87 \times 70) = 0
\]

Let's simplify and solve for \( y_1 \).

### Step 1: Simplify the equation
- First, calculate \( 1017.87 + 1017.87 \), which is \( 2035.74 \).
- Then, calculate \( 1017.87 \times 430 = 437681.1 \) and \( 1017.87 \times 70 = 71250.9 \), and sum these: \( 437681.1 + 71250.9 = 508932 \).

Now substitute these values into the equation:

\[
400y_1^2 + 2 \times 7.54 \times 2035.74y_1 - 2 \times 7.54 \times 508932 = 0
\]

### Step 2: Multiply the constants
- \( 2 \times 7.54 \times 2035.74 = 30688.36 \).
- \( 2 \times 7.54 \times 508932 = 7664577.36 \).

Now the equation becomes:

\[
400y_1^2 + 30688.36y_1 - 7664577.36 = 0
\]

### Step 3: Solve using the quadratic formula
The quadratic formula is:

\[
y_1 = \frac{-

400y_1^2+2×7.54×(1017.87+1017.87)y_1-2×7.54×(1017.87×430+1017.87×70)=0

Learn how to solve the quadratic equation 400y_1^2 + 2 × 7.54 × (1017.87 + 1017.87)y_1 - 2 × 7.54 × (1017.87 × 430 + 1017.87 × 70) with detailed step-by-step simplification. This problem covers algebraic concepts, including quadratic equations, simplification, and the quadratic formula theorem. Suitable for Grades 9-12 students.

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