## anonymous 4 years ago Which expression gives the solutions of –5 + 2x² = –6x?

1. myininaya

2. myininaya

$2x^2+6x-5=0$

3. myininaya

Now use the quadratic formula! :)

4. myininaya

a=2,b=6,c=-5

5. anonymous

6. myininaya

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

7. anonymous

i dont know how to use that. D:

8. myininaya

Do you know how to complete the square?

9. anonymous

no. :/ im horrible at math.

10. myininaya

hmmm... so what method do you want to use?

11. anonymous

i dont know any methods to solve that.

12. ash2326

13. anonymous

okay.

14. ash2326

A standard quadratic equation is given as $ax^2+bx+c=0$ its roots or its solution is given by the following formula $x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

15. ash2326

This formula gives us two values of x which satisfies the quadratic equation

16. ash2326

Did you understand till now?

17. anonymous

formulas confuse me.

18. ash2326

what's confusing you?

19. anonymous

the formula i dont know how to use it.

20. ash2326

It's very easy, I'll show you your question is -5+2x^2=6x The first step is to bring all the terms to one side $-5+2x^2=6x$ Let's subtract both sides by 6x we get $-5+2x^2-6x=6x-6x$ we'll get $-5+2x^2-6x=0$ Now let's write this in descending powers of x $2x^2-6x-5$ Did you understand till now?

21. anonymous

yes. i think i understand that.

22. ash2326

Great now let's compare this with standard equation $ax^2+bx+c=0$ $2x^2-6x+5=0$ we have $a=2,\ b=-6\ and\ c=5$ Did you get this?

23. anonymous

kinda

24. ash2326

Now its solution is given by $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$ Let's substitute a=2, b=-6 and c=-5 $x=\frac{-(-6)\pm \sqrt{(-6)^2-4\times (2)\times (-5)}}{2\times (2)}$ We get now $x=\frac{6\pm \sqrt{36-(-40)}}{4}$ $x=\frac{6\pm \sqrt{36+40}}{4}$ $x=\frac{6\pm \sqrt{76}}{4}$ Did you understand this?

25. anonymous

yes

26. ash2326

We've almost found the solution, we could simplify it more We have $x=\frac{6\pm \sqrt{76}}{4}$ $\sqrt {76}= \sqrt{4\times 19}=2\sqrt {19}$ so $x=\frac{6\pm 2\sqrt{19}}{4}$ divide the whole equation by 2 we get $x=\frac{3\pm \sqrt {19}}{2}$ so the two solutions\roots are $x=\frac{3+ \sqrt {19}}{2}, \frac{3- \sqrt {19}}{2}$\] Did you understand?

27. anonymous

yeah.

28. ash2326

So whenever you've to find roots of quadratic equation, use this formula:D

29. anonymous

okay!