
Using u = 2x + 3
![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}=(-8\pm4.4721)/(2)](https://img.qammunity.org/2023/formulas/mathematics/college/km2639fj3epjsoonz9shho36e6hrb11w6n.png)
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