11-10=10q+10q^2

Solution for 11-10=10q+10q^2 equation:

11-10=10q+10q^2

We move all terms to the left:

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

We add all the numbers together, and all the variables

-(10q+10q^2)+1=0

We get rid of parentheses

-10q^2-10q+1=0

a = -10; b = -10; c = +1;
Δ = b2-4ac
Δ = -102-4·(-10)·1
Δ = 140
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{140}=\sqrt{4*35}=\sqrt{4}*\sqrt{35}=2\sqrt{35}$
$q_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{35}}{2*-10}=\frac{10-2\sqrt{35}}{-20}
$
$q_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{35}}{2*-10}=\frac{10+2\sqrt{35}}{-20}
$