XIII SOLVING QUADRATIC EQUATIONS USING THE QUADRATIC
FORMULA.
A formula for solving quadratic equations can be derived by completing the
square for the standard form of a quadratic equation, 0  ax 2  bx  c . The solutions of
the equation are found by applying the formula :
b  b 2  4ac
x
2a
In the formula, a if the coefficient of the quadratic term ( x 2 ) , b is the coefficient
of the linear term (x), and c is the constant term. The quadratic formula will always yield
the solution to a given quadratic equation, but should be used only when the equation
cannot be factored quickly, or when completing the square is not efficient.
Example 1. Solve 3x 2  2 x  4  0 using the quadratic formula.
In this equation, a = 3, b = 2, and c = -4. Substituting these values
into the formula will result in the solutions to the equation.
x
x
x
2 
22  4(3)(4)
2(3)
2 
4  4  12 
2 
6
4  48
6
2  52
x
6
2  2 13
x
6
2(1  13)
x
2(3)
x
1  13
3
The solution set is 
1  13 1  13
,

3
3
TO SOLVE A QUADRATIC EQUATION USING THE QUADRATIC FORMULA :
1)
2)
3)
4)
Write the equation in standard form, ax 2  bx  c  0
Identify the values of a, b and c in this equation
Replace these values into the quadratic formula
Simplify
Example 2. Solve 2t 2  3t  5  0 over the set of complex numbers.
1) The equation is in standard form.
2) a = 2, b = -3, and c = 5
3) Replace these values in the formula :
x
(3) 
(3) 2  4 (2)(5)
2(2)
3  9  4 (10)
4
3  9  40
x
4
3  31
x
4
3  i 31
x
4
x
The solution set is 
3  i 31 3  i 31
,

4
4