Question

Solve by Completing the square.
2y^2+4y=5

Answer 1

2y^2+4y=5
(y+2)^2 = (y^2 + 4y + 4) - 9 = 2y^2+4y-5=0
=> (y+2)^2 - 9

Answer 2

ummmm can we just check \[(y +2)^2 - 9 = y^2 + 4y + 4 - 9 = y^2 + 4y - 5 ~~which ~~\neq~~2y^2 + 4y - 5\]

Answer 3

\[2y^2 +4y -5=0\]\[2(y^2+2y) - 5 =0\]\[2(y^2+2y+2) - 5 -2=0\]\[2(y+1)^2 -7=0\]\[2(y+1)^2 =7\]\[(y+1)^2 = \frac{7}{2}\]\[y+1=\pm\sqrt{\frac{7}{2}}\]\[y= -1 \pm \sqrt{\frac{7}{2}}\]

Answer 4

2*(y+1)^2 = 2*(y^2 + 2y + 1) = 2y^2 + 4y + 2, so
2(y+1)^2 - 7 = 2y^2 + 4y + 2 -7
2(y+1)^2 - 7 = 2y^2 + 4y - 5

Answer 5

My first answer is wrong, i forgot the 2 multiplying the y^2...