Solving Quadratic Equations Using the Quadratic Formula

Introduction

The goal is to solve quadratic equations of the form ( ax^2 + bx + c = 0 ) using the quadratic formula.
Quadratic formula: [ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

Identify coefficients:
- ( a = 2 )
- ( b = 3 )
- ( c = -2 )
Substitute into the quadratic formula:
- ( x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot (-2)}}{2 \cdot 2} )
- Calculation steps:
  - ( 3^2 = 9 )
  - ( -4 \cdot 2 \cdot (-2) = 16 )
  - ( 9 + 16 = 25 )
  - ( \sqrt{25} = 5 )
- Resulting expressions:
  - ( x = \frac{-3 + 5}{4} ) and ( x = \frac{-3 - 5}{4} )
Simplify solutions:
- ( x = \frac{2}{4} = \frac{1}{2} )
- ( x = \frac{-8}{4} = -2 )
Solutions: ( x = \frac{1}{2} ) and ( x = -2 )
Verification:
- Plug ( x = -2 ) back into the equation:
  - ( 2(-2)^2 + 3(-2) - 2 = 0 ) verifies the solution.

Identify coefficients:
- ( a = 6 )
- ( b = -17 )
- ( c = 12 )
Substitute into the quadratic formula:
- ( x = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 \cdot 6 \cdot 12}}{2 \cdot 6} )
- Calculation steps:
  - ( (-17)^2 = 289 )
  - ( -4 \cdot 6 \cdot 12 = -288 )
  - ( 289 - 288 = 1 )
  - ( \sqrt{1} = 1 )
- Resulting expressions:
  - ( x = \frac{17 + 1}{12} ) and ( x = \frac{17 - 1}{12} )
Simplify solutions:
- ( x = \frac{18}{12} = \frac{3}{2} )
- ( x = \frac{16}{12} = \frac{4}{3} )
Solutions: ( x = \frac{3}{2} ) and ( x = \frac{4}{3} )