Respuesta :

[tex](2x+3)^2+8(2x+3)\text{ + 11 = 0}[/tex]

Using u = 2x + 3

[tex]u^2+8u+11\text{ = 0 }\Longrightarrow\text{ u = }\frac{-8\pm\sqrt[]{8^2-4(1)(11)}}{2(1)}=\frac{-8\pm\sqrt[]{64-44}}{2}=\frac{-8\pm\sqrt[]{20}}{2}=\frac{-8\pm4.4721}{2}[/tex]

Two solutions:

u1 = (-8 + 4.4721)/2 = −1.76395

u2 = (-8 - 4.4721)/2 = −6.23605

x1 ==> 2x + 3 = −1.76395 ==> 2x = −1.76395 - 3 = -4.76395 ==> x1 = -4.76395/2 = −2.381975

x2 ==> 2x + 3 = −6.23605 ==> 2x = −6.23605 - 3 = -9.23605 ==> x2 = -9.23605/2 = −4.618025

Answers:

x1 = −2.381975

x2 = −4.618025

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