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## anonymous one year ago solve the equation using quadratic formula. 2x (sqaured) +13x=0

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1. anonymous

can some one please help ,e i dont get quadratic functions

2. pooja195

Find a GCF first do you see any?

3. anonymous

13?

4. AbdullahM

It says to solve by using the Quadratic Formula... What is the Quadratic Formula?

5. anonymous

"ax2 + bx + c = 0"

6. pooja195

Thats standered form

7. AbdullahM

|dw:1434816541031:dw|

8. anonymous

yeah but for this question thers no c

9. pooja195

so that would be 0 then :)

10. anonymous

but dont u have tp plug in the numbers? or is it when theres no c then the answer is equivalent to 0?

11. pooja195

Name your abc values

12. anonymous

a=2 b=13

13. pooja195

$\huge~\rm~ax^2+bx+c=0$ $\huge~\rm~2x^2+13x+0=0$

14. pooja195

Now name the abc values

15. anonymous

a=2 b=13 c=0?

16. AbdullahM

Great job so far :)

17. pooja195

yes! :) $\huge~\rm~~x=\frac{ -(13)~\pm \sqrt{(13)^2-4(2)(0)} }{ 2(2)}$

18. anonymous

$so.... x= -13 \pm \sqrt{169}$

19. anonymous

and then i would subtract 8?

20. anonymous

i dont understnd now...

21. pooja195

$\huge~\rm~~x=\frac{ -13~\pm \sqrt{169~x~1} }{ 4}$ $\huge~\rm~~x=\frac{ -13~\pm \sqrt{169} }{ 4}$ Whats the square root of 169?

22. anonymous

13

23. anonymous

right?

24. pooja195

$\huge~\rm~~x=\frac{ -13~\pm~13 \sqrt{1} }{ 4}$ now set up two equations and solve $\huge~\rm~~x=\frac{ -13~+~13 \sqrt{1} }{ 4}$ $\huge~\rm~~x=\frac{ -13~-~13 \sqrt{1} }{ 4}$

25. anonymous

so it would be -13/2 and 0?

26. anonymous

those are my two answers correct?

27. anonymous

so would i do the same thing for 6x (sqaured) +2x=4?

28. pooja195

You would subtract the 4 $\huge~\rm~6x^2+2x-4=0$ and go from that

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