Question

Solve x2 + 7x = 1

Answer 1

oh look just another quadratic equation

Answer 2

this one does not factor, so you really do have to use the quadratic formula
subtract \(1\) to get
\[x^2+7x-1=0\] then find
\[\frac{-b\pm\sqrt{b^2-4ac}}{2a}\] with
\[a=1,b=7,c=-1\]

Answer 3

you want to try it yourself?
you still need help?
you just want the answer?

Answer 4

yeah yeah yeah ))):

Answer 5

of course i don't just want the answer hhaaha, ill try

Answer 6

k i will check if you like
gonna go get a peach or something

Answer 7

-7 +/- sqrt 45/2

Answer 8

is how far i got

Answer 9

close

Answer 10

im not sure how to finish it

Answer 11

well it would in fact be finished there but you made a very common sign error
lets do that part right

Answer 12

oh is the 7 positive

Answer 13

inside the radical is \(b^2-4ac\)
in your case \(a=1,b=7,c=-1\) note that \(c=-1\) not \(1\)

Answer 14

no the \(-7\) part is right

Answer 15

ohh

Answer 16

so -2

Answer 17

hold the phone you were closer before
\[b^2-7ac\\
7^2-4\times 1\times (-1)\\
49+4\\
53\]

Answer 18

ohhhhh

Answer 19

you had \(49-4=45\) but it should have been \(49+4\)

Answer 20

ok i understand now

Answer 21

so just change the 45 to 53 and badaboom i got an answer

Answer 22

final answer is
\[\frac{-7\pm\sqrt{53}}{2}\] and as you said badaboom you get the "final answer"
there is nothing you can do with this , leave it

Answer 23

alright awesome !

Answer 24

invoice is in the mail

Answer 25

\(x^2 + 7x = 1\)

\(x^2 + 7x + \left(\dfrac{7}{2}\right)^2= 1 + \left(\dfrac{7}{2}\right)^2\)

\(\left(x + \dfrac{7}{2}\right)^2= 1 + \dfrac{49}{4}\)

\(\left(x + \dfrac{7}{2}\right)^2= \dfrac{4}{4} + \dfrac{49}{4}\)

\(\left(x + \dfrac{7}{2}\right)^2= \dfrac{53}{4}\)

\(x + \dfrac{7}{2}= \pm \sqrt{\dfrac{53}{4}}\)

\(x = -\dfrac{7}{2} \pm \dfrac{\sqrt{53}}{\sqrt{4}}\)

\(x = -\dfrac{7}{2} \pm \dfrac{\sqrt{53}}{2}\)

\(x = -\dfrac{7 \pm \sqrt{53}}{2}\)