## mathslover 3 years ago Hi friends this is a small tutorial regarding the proof of quadratic equation ... this is made by me

1. mathslover

2. terenzreignz

3. mathslover

yes @terenzreignz $\Huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

4. mathslover

${\textbf{the general form of a quadratic equation is ax^2+bx+c=0 . .}}$ $\textbf{it is known as quadratic equation since it's degree is 2 }$ HOW TO GET THE ROOT OF A QUADRATIC EQUATION : $\huge{ax^2+bx+c=0}$ $\huge{ax^2+bx=-c}$ $\huge{\frac{(ax^2+bx)}{a}=\frac{-c}{a}}$ $\huge{x^2+\frac{bx}{a}=\frac{-c}{a}}$ $\huge{(x)^2+2(\frac{b}{2a})(x)=\frac{-c}{a}}$ $\huge{(x)^2+2(\frac{b}{2a})(x)+\frac{b^2}{4a^2}=\frac{-c}{a}+\frac{b^2}{4a^2}}$ $\huge{(x+\frac{b}{2a})^2 = \frac{-c}{a}+\frac{b^2}{4a^2}}$ $\huge{x+\frac{b}{2a}=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}}$ $\huge{x=\pm\sqrt{\frac{-c}{a}+\frac{b^2}{4a^2}}-\frac{b}{2a}}$ $\huge{x=\pm\sqrt{\frac{b^2-4ac}{4a^2}}-\frac{b}{2a}}$ $\huge{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}$

5. ParthKohli

Good work, mathslover, I appreciate your tutorial :)

6. mathslover

thanks a lot @ParthKohli

7. mathslover

Please prefer the LaTeX for better proof and for no mistakes

Great job @mathslover!

9. mathslover

10. mathslover

You are very welcome

12. Yahoo!

@mathslover wat is ur intention by posting this!

13. mathslover

to just make you all aware with quadratic equation

14. Yahoo!

ok

15. mathslover

16. Yahoo!

17. mathslover

i will but after some time. .. doing some research work :)

18. wdkekefkjrgjr

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19. mathslover

20. mathslover

i m posting more interesting things about quadratic equation as a new question

21. mathslover
22. jiteshmeghwal9

k! thanx a lot @mathslover :)

23. mathslover

Ur welcome @jiteshmeghwal9 Best of Luck