Question

solve by completing the square. 4x^2+4x+3=0

Answer 1

4x^2+4x+3=0

you'll get
a = 4
b = 4
c = 3

subsitute all of them (a, b, and c) to quadratic equations

Answer 2

yes do that lolz

Answer 3

\[x = \frac{ -b \pm \sqrt{b^2 -4ac} }{ 2a } \]

Answer 4

what else can do

Answer 5

Check your equations

Is your equations

4x^2-4x+3=0
4x^2+4x-3=0
or
4x^2-4x-3=0
????

Answer 6

4x^2+4x+3=0

Answer 7

ok

Answer 8

\[x = \frac{ -b \pm \sqrt{b^2 - 4*a*c} }{ 2a } = \frac{ -4 \pm \sqrt{4^2 - 4*4*3} }{ 2*4 } = \frac{ -4 \pm \sqrt{-32} }{ 8 }\]

Answer 9

please help. solve by completing the square 4x^2+4x+3=0

Answer 10

\[x = \frac{ -4 \pm \sqrt{-1}*\sqrt{32} }{ 8} = \frac{ -4 \pm i*\sqrt{32} }{ 8 } = \frac{ -4 \pm 4\sqrt{2}*i }{ 8 }\]

Answer 11

so you will get
\[x = \frac{ -4 + 4\sqrt{2}*i }{ 8 } =\frac{ -1 + \sqrt{2}*i }{ 2 } = -\frac{ 1 }{ 2 } + \frac{ 1 }{ 2 } \sqrt{2} *i\]

Answer 12

and

Answer 13

\[x = \frac{ -4 - 4\sqrt{2} *i}{ 8 } = \frac{ -1 - \sqrt{2}*i }{ 2} = -\frac{ 1 }{ 2 } - \frac{ 1 }{ 2 } \sqrt{2}*i\]

Answer 14

i did it for the lulz
i did it for the lulz
i did it for the lul

Answer 15

get it @eogbonna ??

Answer 16

help with solve by completing the square 4x^2+4x+3=0

Answer 17

help ,i am running out of time