Question

Algebra question

Answer 1

\[9x^2+3x-2=0\]

Answer 2

hey, have you heard of Discriminant?

Answer 3

im trying to solve this but a got stuck on this step : \[x= (3\pm \sqrt{81})/18\]

Answer 4

you know that \[\sqrt{81}=\pm9\] right?:)

Answer 5

You are doing everything correctly

Answer 6

this is all my work: \[x=\frac{ -{3\pm \sqrt{3^2-4(9)(-2)} } }{ 18}\]

Answer 7

\[x=\frac{ -3\pm \sqrt{9+72} } {18} \]

Answer 8

\[\frac{ -3\pm \sqrt {81} }{ 18 }\]
\[x=\frac{ -3\pm9 }{ 18 }\]

Answer 9

and thats it i get stuck from there

Answer 10

well, +/- means that there are x1 and x2, so x1=(-3+9)/18=6/18=1/3 and x2=(-3-9)/18=-12/18=-2/3

Answer 11

did you understand or want me to write in latex?

Answer 12

i prefer you write it in latex^^

Answer 13

please

Answer 14

So,the main formula for solutions of quadratic equation \[x _{1,2}=\frac{ -b \pm \sqrt{b ^{2}-4ac} }{ 2a }\] where \[x _{1,2}\] means that there are two solutions
\[x _{1}\] and \[x _{2}\]

so to find \[x _{1}\]:
\[x _{1}=\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a}\]
and to find \[x _{2}\]
\[x _{2}=\frac{ -b-\sqrt{b ^{2}-4ac} }{ 2a }\]

so now according to your problem:
\[x _{1}=\frac{ -b+\sqrt{b ^{2}-4ac} }{ 2a }=\frac{ -3 +\sqrt{3^{2}+4*(-2)*9}}{ 2*9 }=\frac{ -3+\sqrt{81} }{ 18}=\frac{ -3+9 }{ 18 }=\frac{ 6 }{ 18}=\frac{ 1 }{ 3 }\]

Answer 15

sorry Ilost my connection

Answer 16

could you do \[x _{2}\] for me?

Answer 17

no its ok now i got it. thank you so much is just that im taking virtual classes and have noone to help me so sometimes i dont understand some stuff thank you so much O.O

Answer 18

i just wanted someone to explain me more about the quadratic formula

Answer 19

you are welcome! I am glad I could help you