Answer:
[tex]\large\boxed{x^2+10x+5=\left(x+5+2\sqrt5\right)\left(x+5-2\sqrt5\right)}[/tex]
Step-by-step explanation:
[tex]\text{Use the quadratic formula:}\\\\ax^2+bx+c=a\left(x-\dfrac{-b-\sqrt{b^2-4a}}{2a}\right)\left(x-\dfrac{-b+\sqrt{b^2-4ac}}{2a}\right)\\======================================[/tex]
[tex]\text{We have:}\ x^2+10x+5\to a=1,\ b=10\ \text{and}\ c=5\\\\\text{Substitute}\\\\b^2-4ac=10^2-4(1)(5)=100-20=80\\\sqrt{b^2-4ac}=\sqrt{80}=\sqrt{16\cdot5}=\sqrt{16}\cdot\sqrt5=4\sqrt5\\\\x_1=\dfrac{-10-4\sqrt5}{2(1)}=-5-2\sqrt5\\\\x_2=\dfrac{-10+4\sqrt5}{2(1)}=-5+2\sqrt5\\\\\bigg(x-\left(-5-2\sqrt5\right)\bigg)=\left(x+5+2\sqrt5\right)\\\\\bigg(x-\left(-5+2\sqrt5\right)\bigg)=\left(x+5-2\sqrt5\right)[/tex]
