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Line 11: − + + + + + + + + + + + + + + + + + + + + − Derivation of the formula is usually done via the method of [[completing the square]].
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:<math>x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}</math>
:<math>x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}</math>
You can prove that the formula is correct by substituting the formula in place of '''x''' in <math>ax^2+bx+c=0\!</math> and then gradually simplifying the rather complicated formula that results, step by step. Eventually, if all the steps are done correctly, it will simplify to 0.
You can prove the formula the following way:
:<math>ax^2+bx+c=0\!</math>
:<math>x^2+\frac{b}{a}x+\frac{c}{a}=0\!</math>
:<math>(x+\frac{b}{2a})^2-(\frac{b}{2a})^2+\frac{c}{a}=0\!</math>
:<math>(x+\frac{b}{2a})^2=(\frac{b}{2a})^2-\frac{c}{a}\!</math>
:<math>(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2}\!</math>
:<math>x+\frac{b}{2a}=\frac{\pm \sqrt {b^2-4ac}}{2a}\!</math>
:<math>x = \frac{-b \pm \sqrt {b^2-4ac}}{2a}</math>
This method of deriving the formula is done via the method of [[completing the square]].
You can assert that the formula is correct by substituting the formula in place of '''x''' in <math>ax^2+bx+c=0\!</math> and then gradually simplifying the rather complicated formula that results, step by step. Eventually, if all the steps are done correctly, it will simplify to 0.
[[Category:Mathematics]]
[[Category:Mathematics]]