Question

Solve 3x^2 + 18x + 15 = 0 by completing the square.

A: 3(x + 3)^2 = 6; x = −3 + √2, x = −3 − √2
B: 3(x + 9)^2 = 12; x = −11, x = −7
C: 3(x + 3)^2 = 12; x = −5, x = −1
D: 3(x + 9)^2 = 228; x = −9 + √76, x = −9 − √76


My Math:
3x^2 + 18x + 15 = 0
3(x^2 + 18x + ) + 15 = 0
3(x^2 + 18x + 81) + 15 - 81 = 0
(x^2 + 18x + 81) - 228 = 0

What do I do now? Can someone help me please?

Answer 1

c

Explanation:

Answer 2

You made a mistake in your second line.

You put
`3x^(2) +18x` in brackets, dividing the
`3 x^(2)` by 3, but you didn't divide the 18x by 3.

So you ended up with 3 times 18x in the end.

You should divide the 18x by 3 too, so you end up with
`3( x^(2) +6x)+15=0`