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## anonymous one year ago help solving quadratic equation 8x+60=x^2+15x Help please please cx

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1. Nnesha

set it equal to 0 so move 8x+60 to the right side

2. anonymous

so 0=x^2+15x- 8x+60?

3. Nnesha

nope 60 is positive at left side so subtract 60 both sides

4. anonymous

ohhh so 0=x^2+15x- 8x-60?

5. Nnesha

yes right now combine LIKE terms

6. anonymous

0=x^2+15x- 8x-60 ^ 7 0=x^2+7x-60

7. anonymous

@Nnesha what do i do next?

8. Nnesha

yes that's right now you can use quadratic formula to solve for x $\huge\rm \frac{ {- }b \pm \sqrt{b^2 -4ac} }{ {2a} }$ or you can factor it $\large\rm \color{red}{1}x^2 +\color{Red}{ 7}x \color{reD}{- 60} = 0$$\large\rm \color{blue}{A}x^2 +\color{blue}{B}x+\color{blue}{C}=0$

9. anonymous

the factoring one seems simpler lol

10. anonymous

i think thats what im suppose to do, is to factor it

11. Nnesha

alright to factor this one find two number if you multiply them you should get product of AC and if you add or subtract them you should get middle term b

12. Nnesha

A is leading coefficient C is constant term :-)

13. anonymous

-5 and 12

14. Nnesha

yep right so bec leading coefficient is just one GREAT! just write those two numbers with x (x + 1st number)(x+ 2nd number)

15. anonymous

(x+(-5))(x+12)

16. Nnesha

yes right + times -5 equal - so (x-5)(x+12)=0

17. Nnesha

if you have to solve for x set both parentheses equal to zero solve for x like this x-5 =0 and x+12 =0

18. anonymous

ohh ok, i would get 5 and -12

19. Nnesha

yep that's right!

20. anonymous

Thank you!, is that it?

21. anonymous

the answer

22. Nnesha

yep factors are (x-5)(x+12) solution or x value are 5 and -12

23. anonymous

Thank you so much! Cx

24. Nnesha

my pleasure :-)

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