## qudrex Group Title Which expression gives the solutions of –5 + 2x² = –6x? 2 years ago 2 years ago

1. myininaya Group Title

2. myininaya Group Title

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

3. myininaya Group Title

Now use the quadratic formula! :)

4. myininaya Group Title

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

5. qudrex Group Title

6. myininaya Group Title

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

7. qudrex Group Title

i dont know how to use that. D:

8. myininaya Group Title

Do you know how to complete the square?

9. qudrex Group Title

no. :/ im horrible at math.

10. myininaya Group Title

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

11. qudrex Group Title

i dont know any methods to solve that.

12. ash2326 Group Title

13. qudrex Group Title

okay.

14. ash2326 Group Title

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 Group Title

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

16. ash2326 Group Title

Did you understand till now?

17. qudrex Group Title

formulas confuse me.

18. ash2326 Group Title

what's confusing you?

19. qudrex Group Title

the formula i dont know how to use it.

20. ash2326 Group Title

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. qudrex Group Title

yes. i think i understand that.

22. ash2326 Group Title

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. qudrex Group Title

kinda

24. ash2326 Group Title

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. qudrex Group Title

yes

26. ash2326 Group Title

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. qudrex Group Title

yeah.

28. ash2326 Group Title

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

29. qudrex Group Title

okay!