Answer:
[tex]x=-2.38[/tex]
[tex]x=-4.62[/tex]
Step-by-step explanation:
The question is [tex](2x+3)^2+8(2x+3)+11=0[/tex]
We let [tex]u=2x+3[/tex], so the equation becomes:
[tex]u^2+8u+11=0[/tex]
Where [tex]a=1, b=8, c=11[/tex]
Putting it in the quadratic formula, we have:
Quadratic formula: [tex]\frac{-b+-\sqrt{b^2-4ac} }{2a}[/tex]
Substituting we have: [tex]\frac{-8+-\sqrt{(8)^2-4(1)(11)} }{2(1)}\\=\frac{-8+-\sqrt{20} }{2}\\=\frac{-8+-2\sqrt{5} }{2}\\=-4+\sqrt{5}, -4-\sqrt{5} }[/tex]
We let [tex]u=2x+3[/tex], so x is:
[tex]u=2x+3\\(-4+\sqrt{5})=2x+3\\x=\frac{-7+\sqrt{5}}{2}=-2.38[/tex]
and
[tex]u=2x+3\\(-4-\sqrt{5})=2x+3\\x=\frac{-7-\sqrt{5}}{2}=-4.62[/tex]
The solutions of the equation is [tex]x=-2.38[/tex] (rounded to 2 decimal places), and [tex]x=-4.62[/tex] (rounded to 2 decimal places)
