6b+11=b^2

Solution for 6b+11=b^2 equation:

6b+11=b^2

We move all terms to the left:

6b+11-(b^2)=0

determiningTheFunctionDomain
-b^2+6b+11=0

We add all the numbers together, and all the variables

-1b^2+6b+11=0

a = -1; b = 6; c = +11;
Δ = b2-4ac
Δ = 62-4·(-1)·11
Δ = 80
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{80}=\sqrt{16*5}=\sqrt{16}*\sqrt{5}=4\sqrt{5}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-4\sqrt{5}}{2*-1}=\frac{-6-4\sqrt{5}}{-2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+4\sqrt{5}}{2*-1}=\frac{-6+4\sqrt{5}}{-2}
$