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