Answer:
[tex]\huge\boxed{\left\{-\dfrac{5+\sqrt{21}}{2};\ \dfrac{-5+\sqrt{21}}{2}\right\}}[/tex]
Step-by-step explanation:
[tex]x^2+5x+1=0\\\\\text{Use the quadratic formula:}\\\\\text{For}\ ax^2+bx+c=0\\\\\Delta=b^2-4ac\\\\\text{if}\ \Delta < 0,\ \text{then the equation has no real solution}\\\text{if}\ \Delta=0,\ \text{then the equation has one real solution}\ x=\dfrac{-b}{2a}\\\text{if}\ \Delta>0,\ \text{then the equation has two real solution}\ x=\dfrac{-b\pm\sqrt{\Delta}}{2a}[/tex]
[tex]\text{We have}\\\\a=1,\ b=5,\ c=1\\\\\Delta=5^2-4(1)(1)=25-4=21>0\\\\x=\dfrac{-5\pm\sqrt{21}}{2(1)}=\dfrac{-5\pm\sqrt{21}}{2}[/tex]
