Answer:
[tex]x=\frac{-7-\sqrt{5}}{2}[/tex] or [tex]x=\frac{-7+ \sqrt{5}}{2}[/tex]
Step-by-step explanation:
The given equation is
[tex](2x+3)^2+8(2x+3)+11=0[/tex]
Let us treat this as a quadratic equation in [tex](2x+3)[/tex].
where [tex]a=1,b=8,c=11[/tex]
The solution is given by the quadratic formula;
[tex](2x+3)=\frac{-b\pm \sqrt{b^2-4ac} }{2a}[/tex]
We substitute these values into the formula to obtain;
[tex](2x+3)=\frac{-8\pm \sqrt{8^2-4(1)(11)} }{2(1)}[/tex]
[tex](2x+3)=\frac{-8\pm \sqrt{64-44} }{2}[/tex]
[tex](2x+3)=\frac{-8\pm \sqrt{20} }{2}[/tex]
[tex](2x+3)=\frac{-8\pm2\sqrt{5} }{2}[/tex]
[tex](2x+3)=-4\pm \sqrt{5}[/tex]
[tex](2x+3)=-4-\sqrt{5}[/tex] or [tex](2x+3)=-4+ \sqrt{5}[/tex]
[tex]2x=-3-4-\sqrt{5}[/tex] or [tex]2x=-3-4+ \sqrt{5}[/tex]
[tex]2x=-7-\sqrt{5}[/tex] or [tex]2x=-7+ \sqrt{5}[/tex]
[tex]x=\frac{-7-\sqrt{5}}{2}[/tex] or [tex]x=\frac{-7+ \sqrt{5}}{2}[/tex]
