Solve the equation $\left| { x }^{ 2 }+4x+3 \right| = -\left( 2x+5 \right)$.
+49 \({x}^{2}+4x+3=-(2x+5)\)
\({x}^{2}+4x+3=-2x-5\)
\({x}^{2}+4x+3+5=-2x-5+5\)
\({x}^{2}+4x+8=-2x \)
\(x^2+4x+8+2x=-2x+2x\)
\(x^2+6x+8=0\)
\(x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \:8}}{2\cdot \:1}\)
\(\sqrt{6^2-4\cdot \:1\cdot \:8}=2 \)
\(x_{1,\:2}=\frac{-6\pm \:2}{2\cdot \:1}\)
\(x_1=\frac{-6+2}{2\cdot \:1},\:x_2=\frac{-6-2}{2\cdot \:1}\)
x=-2
x=-4
+49 \({x}^{2}+4x+3=-(2x+5)\)
\({x}^{2}+4x+3=-2x-5\)
\({x}^{2}+4x+3+5=-2x-5+5\)
\({x}^{2}+4x+8=-2x \)
\(x^2+4x+8+2x=-2x+2x\)
\(x^2+6x+8=0\)
\(x_{1,\:2}=\frac{-6\pm \sqrt{6^2-4\cdot \:1\cdot \:8}}{2\cdot \:1}\)
\(\sqrt{6^2-4\cdot \:1\cdot \:8}=2 \)
\(x_{1,\:2}=\frac{-6\pm \:2}{2\cdot \:1}\)
\(x_1=\frac{-6+2}{2\cdot \:1},\:x_2=\frac{-6-2}{2\cdot \:1}\)
x=-2
x=-4
+267 @IWantToHelp: There is an edit feature on web2.0calc. If you want to add on to you post just edit it. Instead of posting twice just edit your first post (if you want.)
