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Quadratic Equation One Unknown

Reference data and engineering information about quadratic equation one unknown for miscellaneous applications.

quadraticequationoneunknown

Overview

Engineering reference data for Quadratic Equation One Unknown in miscellaneous.

Key Formulas

Unit Conversion

y=x⋅ky = x \cdot k

Multiply by conversion factor.

Linear Interpolation

y=y1+(x−x1)(y2−y1)x2−x1y = y_1 + \frac{(x - x_1)(y_2 - y_1)}{x_2 - x_1}

Estimate between two known points.

Percentage

p=partwhole×100%p = \frac{\text{part}}{\text{whole}} \times 100\%

Part as fraction of whole.

Variables

Symbol Description Unit
xx Input value —
yy Output value —
kk Conversion factor —

Example Calculation

The following demonstrates solving a quadratic equation for specific coefficients.

Given:
a=1,b=6,c=5a = 1, \quad b = 6, \quad c = 5

Step 1: Substitute into the quadratic formula.
x1,2=−b±b2−4ac2ax_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

Step 2: Compute the values.
For the first root:
x1=−6+62−4⋅1⋅52⋅1=−6+162=−6+42=−1x_1 = \frac{-6 + \sqrt{6^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{-6 + \sqrt{16}}{2} = \frac{-6 + 4}{2} = -1

For the second root:
x2=−6−62−4⋅1⋅52⋅1=−6−162=−6−42=−5x_2 = \frac{-6 - \sqrt{6^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{-6 - \sqrt{16}}{2} = \frac{-6 - 4}{2} = -5

Result:
The solutions to the equation x2+6x+5=0x^2 + 6x + 5 = 0 are x1=−1x_1 = -1 and x2=−5x_2 = -5.

References