Question 322195: Please show me how to solve a quadratic equation using the quadratic formula. Can you please provide an example of a quadratic equation and show how you solved it? I am struggling with this topic.
Answer by solver91311(24713)   (Show Source):
You can put this solution on YOUR website!
The general solution of a quadratic equation in standard form, namely  is:
Example:
Here,  ,  , and 
\ \pm\ \sqrt{(-2)^2\ -\ 4(1)(-3)}}{2(1)}\ =\ \frac{2\ \pm\ \sqrt{4\ +\ 12}}{2}\ =\ \frac{2\ \pm\ 4}{2}\ =\ 1\ \pm\ 2)
Therefore,  or 
Example 2:
Here, ,  , and 
\ \pm\ \sqrt{(-1)^2\ -\ 4(1)(-1)}}{2(1)}\ =\ \frac{1\ \pm\ \sqrt{1\ +\ 4}}{2}\ =\ \frac{1\ \pm\ \sqrt{5}}{2})
Therefore,  or 
Example 3:
Here, , , and 
\ \pm\ \sqrt{(-1)^2\ -\ 4(1)(4)}}{2(1)}\ =\ \frac{1\ \pm\ \sqrt{1\ -\ 16}}{2}\ =\ \frac{1\ \pm\ \sqrt{-15}}{2}\ =\ \frac{1\ \pm\ i\sqrt{15}}{2})
Therefore,  or 
Where  is the imaginary number defined by 
Where did it all come from?
Consider the general form of the quadratic:
Add the additive inverse of the constant term to both sides.
Multiply both sides by the reciprocal of the lead coefficient:
Divide the first degree coefficient by 2, square the result, then add that result to both sides:
Factor the perfect square LHS and combine the fractions in the RHS:
^2\ =\ \frac{b^2\ -\ 4ac}{4a^2})
Take the square root of both sides, considering both positive and negative roots:
Add  to both sides:
John
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