How do you solve $x^2+4x+3=0$ ?

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Answer 1 · 2016-03-11T12:06:34

$x=-1and x=-3$

$=>x^2+3x+x+3$
$=>x(x+3)+1(x+3)$
$=>(x+3)(x+1)$
$x=-1and x=-3$

Answer 2 · 2016-03-11T13:03:28

x = -1 ; x = -3

Since (a - b + c) = 0, use shortcut --> x = -1 and $x = -c/a = -3$ .

Reminder of Shortcut.
- When a + b + c = 0 --> x = 1 and $x = c/a$
- When a - b + c = 0 --> x = - 1 and $x = - c/a$

How do you solve x^2+4x+3=0x^2+4x+3=0?