Question

3x^2+5x+4=0

Answer 1

formula

Answer 2

-5+-square root of -23/6

Answer 3

?

Answer 4

oh let me check

Answer 5

no no zeros

Answer 6

here
\[b^2-4ac=-23\]

Answer 7

oh i see you wrote
\[\frac{-5\pm\sqrt{-23}}{6}\]

Answer 8

i got the answer correct it was 5+-i square root of 23/9

Answer 9

ok fine if you are working with complex numbers this is ok. for real numbers no such thing as
\[\sqrt{-23}\]

Answer 10

yes that is right

Answer 11

oops /6

Answer 12

so i got another problem 1/8y^2+y+2=0

so i multiplied by 8 and got y^2+8y+16=0

now when i complete the square i for (y+4)^2

Answer 13

you get
\[(y+4)^2=-16+16=0\]

Answer 14

which is another way of saying
\[x^2+8x+16\] is a perfect square and so
\[x^2+8x+16=0\]
\[(x+4)^2=0\]
\[x=-4\]