## whadduptori Group Title Use the quadratic formula to solve the equation? 4x2 – 6x – 10 = 0 is anybody able to show me work? one year ago one year ago

1. Rezz5 Group Title

the quadratic formula is -b (±)Sqrt[b-4ac]/2a$-b \pm \frac{ \sqrt{b-4ac}}{ 2a}$

2. Rezz5 Group Title

where ax^2 +bx +c just substitute and you will get your answer.

3. jim_thompson5910 Group Title

$\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$ $\Large x = \frac{-(-6)\pm\sqrt{(-6)^2-4(4)(-10)}}{2(4)}$ $\Large x = \frac{6\pm\sqrt{36-(-160)}}{8}$ I'll let you finish.

4. whadduptori Group Title

I have all that work! I just get confused with the last part? like would i subtract 36-(-160)

5. Rezz5 Group Title

36--160? well double negative is positive, so? 36+160

6. jim_thompson5910 Group Title

36-(-160) is the same as 36+160

7. whadduptori Group Title

so it would become 196?

8. jim_thompson5910 Group Title

it would

9. whadduptori Group Title

|dw:1352933217966:dw|

10. whadduptori Group Title

so would that be the answer?

11. jim_thompson5910 Group Title

you can keep going and there's a +- between the 6 and square root

12. jim_thompson5910 Group Title

you should have $\Large x = \frac{6\pm\sqrt{196}}{8}$ but you can keep going

13. Rezz5 Group Title

(6*14)/8 =(3*14)/4 =(3*7)/2 =21/2

14. whadduptori Group Title

yes i have that!

15. Rezz5 Group Title

okay, what is wrong?

16. whadduptori Group Title

wait where did you get the 6*14/8

17. jim_thompson5910 Group Title

the square root of 196 is 14, so... $\Large x = \frac{6\pm\sqrt{196}}{8}$ becomes $\Large x = \frac{6\pm14}{8}$

18. Rezz5 Group Title

above is very clear and correct, so first you do the plus sign which is (6+14)/8 this is ONE OF YOUR x's the you do (6-14)/8 this is your other x value so your x values are: 5/2 and -1

19. whadduptori Group Title

THank you so much! I didnt know how to find that! but you just made it clear for me! thank you both for your help! (:

20. Rezz5 Group Title