## dddan Group Title if a=2 ; b=-5; c=5 how would you solve the quadratic equation? one year ago one year ago

1. hartnn Group Title

then the two roots of x are: $$\huge{x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}$$

2. wio Group Title

Use the quadratic foumula $x_1 =\frac{ -b + \sqrt{b^2-4ac}} {2a}$ $x_2 =\frac{ -b - \sqrt{b^2-4ac}} {2a}$

3. dddan Group Title

ya but i got 4+- the square root of -24 / 4 does this make sense?

4. hartnn Group Title

your b=-5, how did u get 4+/- ... ?

5. dddan Group Title

sry the b=-4 and the c=5

6. wio Group Title

You have imaginary roots. Is that what you're worried about?

7. dddan Group Title

yup

8. wio Group Title

Why? Is it a physics problem and you are supposed to have real roots? Because there is nothing wrong with the quadratic formula. Double check your coefficients

9. hartnn Group Title

u actually have imaginary roots for this..

10. dddan Group Title

alright can u tell me where to take it from the imaginary roots?

11. hartnn Group Title

how did u get square root of -24 ?

12. dddan Group Title

because with the coefficients, its (-4)^2-4(2)(5) , in the square root right?

13. hartnn Group Title

lol! b=-5 not -4

14. dddan Group Title

no i i mixed up the coefficents in the original question. i was given a=2;b=-4;c=5

15. hartnn Group Title

ok, $$\huge{x=\frac{4 \pm \sqrt{-24}}{4}}=\frac{4 \pm 2\sqrt{-6}}{4}=\frac{2\pm i\sqrt{6}}{2}$$

16. dddan Group Title

alright thank you very much!