11=a^2+10a+0

Solution for 11=a^2+10a+0 equation:

11=a^2+10a+0

We move all terms to the left:

11-(a^2+10a+0)=0

We get rid of parentheses

-a^2-10a-0+11=0

We add all the numbers together, and all the variables

-1a^2-10a+11=0

a = -1; b = -10; c = +11;
Δ = b2-4ac
Δ = -102-4·(-1)·11
Δ = 144
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{144}=12$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-12}{2*-1}=\frac{-2}{-2}
=1
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+12}{2*-1}=\frac{22}{-2}
=-11
$