Question

7x^2+9x+8=0

Answer 1

Compare your quadratic equation with \(ax^2+bx+c=0\)
find \[a=...?\\b=...?\\c=...?\\\] \[ \\ \sqrt{b^2-4ac}=...?\]
then the two roots of x are:
\(\huge{x_{1,2}=\frac{-b \pm \sqrt{b^2-4ac}}{2a}}\)

Answer 2

ok im stuck I have up to -9+- sq root 9^2 - 448 / 14

Answer 3

Carry on.

Answer 4

whats 9^2 - 448 =... ?

Answer 5

wait, how did u get 448 ?

Answer 6

oh hold on

Answer 7

4ac = 4*7*8 = ...

Answer 8

ok

Answer 9

ok -143 so how is that then -9+_i sq root 143 /14

Answer 10

yes!
let me type that nicely,
\(\huge x = \dfrac{-9 \pm i\sqrt{143}}{14}\)

Answer 11

thank you

Answer 12

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