We have `|x^(2)+4x+3|+2x+5=0`
or `|x^(2)+4x+3|=-(2x+5)`
Roots of equation are values of x where the graphs of `y=|x^(2)+4x+3|andy=-(2x+5)`
From the graph, one root `xlt-3` and one root for `-3ltxlt-1`.
For `xlt-3`,
`x^(2)+4x+3=-2x-5`
or `x^(2)+6x+8=0`
`implies(x+4)(x+2)=0`
`impliesx=-4(asxlt-3)`
For `-3ltxlt-1`,
`-(x^(2)+4x+3)=-(2x+5)`
`impliesx^(2)+2x-2=0`
`impliesx=(-2pmsqrt(4+8)=-1sqrt3(as-3ltxlt-1)`
Hence `x=-4,-1-sqrt3`
