Question

solve. y2 − y + 5 = 0

Answer 1

hint: use quadratic formula \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
does that help?

Answer 2

nah not really

Answer 3

do you know what the quadratic formula is?

Answer 4

yeah my teacher told us but he just gave us the homework and expected us to do this and turn it in tomorrow

Answer 5

do you know how to use the quadratic formula?

Answer 6

nope my first time

Answer 7

but you have some basic background on it right? like the quadratic equation is in the form \[ax^2 + bx + c = 0\]

Answer 8

somewhat lol

Answer 9

okay for example you have \[x^2 + 2x + 3 = 0\]
in this case a is 1 because the coefficient of x^2 here is 1
b is 2 because the coefficient of x us 2
and c is 3 because the constant is 3

so if i substitute a = 1; b = 2; c = 3

into \[x = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}\]
\[ \implies x = \frac{(-2) \pm \sqrt{(2)^2 - 4(3)(1)}}{2(1)}\]
\[\implies x = \frac{-2 \pm \sqrt{4 - 12}}{2}\]
\[\implies x = \frac{-2 \pm \sqrt{-8}}{2}\]
does that help?

Answer 10

Somewhat... I'll try this out and see if I get the answer

Answer 11

go for it

Answer 12

go to mathway.com and type problem in

Answer 13

it just shows the answer I need the step by step

Answer 14

have you tried using the quadratic formula yet?

Answer 15

yes sir, it's not working for me smh

Answer 16

Math is my weakest subject

Answer 17

lol i know your pain =))

Answer 18

okay here's a hint..

in y^2 - y + 5

a = 1
b = -1
c = 5

does that help?

Answer 19

so I'm writing y as a factor

Answer 20

if you look at the graph of your quadratic equation, you will notice that the function does not cross the x axis at all. That means that you have a complex solution

Answer 21

you will not be able to factor this one at all

Answer 22

now I see why I hate math