Question

Solve using the quadratic formula
2x^2-5x-10=0

Answer 1

use the quadratic formula and sub a=2, b=-5 and c=-10 into it.

Answer 2

i dont know how to do quadratic formula

Answer 3

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

Answer 4

\[ \text{Quadratic Formula:}~~x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
You pretty much only need to substitute the values and simplify.

Answer 5

\[ x=\frac{-b \pm \sqrt {b^2-4ac}}{2a}, \]

Answer 6

yup, @AccessDenied is right and remember to SIMPLlFY!!

Answer 7

so then it would be -5 +_ 5 over 2?

Answer 8

okay im not getting a correct anwser...im getting -105 over 4....

Answer 9

is it -5 pulse or minus sqrt of -105 over 4?

Answer 10

ur answer should be right. It should be +/- Sqrt 105 over 4

Answer 11

its telling me the 105 should be positive....idk ..

Answer 12

\[ \frac{-(-5) \pm \sqrt{(-5)^2 - 4(2)(-10)}}{2(2)} \]
That's how it should look with substitutions
The 105 is correct, but I don't think its negative since (-5)^2 = +25 and -4*2*(-10) = +80

Answer 13

ur answer should be right. It should be +/- Sqrt 105 plus five** over 4

Answer 14

so then the five in front is positive?

Answer 15

yes indeed.

Answer 16

-1.25 + 1.854049621i, -1.25 - 1.854049621i

Answer 17

No imaginary roots, \[ b^2 - 4ac = 25 - 4(2)(-10) = 25 + 80 ~~\text{which is positive, => real roots} \]

Answer 18

@AccessDenied i have another question like that but the anwsers are not the same. the equation is 3x^2-8x+5=0.
The anwsers for this is none of the above, x=5 x=3, x=5/3 x=1, or 8pulse or minus sqrt 4 over 2. I worked it out and dont get any of those anwsers

Answer 19

\[ \begin{split} \frac{-(-8) \pm \sqrt{(-8)^2 - 4(3)(5)}}{2(3)} &= \frac{8 \pm \sqrt{64 - 60}}{6}\\
&= \frac{8 \pm \cancel{\sqrt{4}}2}{6} \end{split}\]

Answer 20

so what would the anwser be? thats no one of the anwser

Answer 21

would it be 10/6 or simplified 5/3 or 6/6 simplified 1

Answer 22

(8+2)/(6) = 10/6 = 5/3, and (8-2)/(6) = 6/6 = 1
\( \pm \) just meaning that we're adding for one case and subtracting for a second case, we would evaluate for both cases to get the two solutions

Answer 23

okay i get it now thank you !!

Answer 24

\[ x = \frac{ -\color{red}b \pm \sqrt{\color{red}{b}^2 - 4\color{blue}a \color{green}c}}{2\color{blue}a}~~~\text{Given}~~\color{blue}ax^2 + \color{red}bx + \color{green}c = 0 \]
Basically for quadratic equation method, you really only need the formula and to be careful of all the sign changing that can occur in the problem. You're welcome! :)