9b+10=110-b^2

Solution for 9b+10=110-b^2 equation:

9b+10=110-b^2

We move all terms to the left:

9b+10-(110-b^2)=0

We get rid of parentheses

b^2+9b-110+10=0

We add all the numbers together, and all the variables

b^2+9b-100=0

a = 1; b = 9; c = -100;
Δ = b2-4ac
Δ = 92-4·1·(-100)
Δ = 481
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(9)-\sqrt{481}}{2*1}=\frac{-9-\sqrt{481}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(9)+\sqrt{481}}{2*1}=\frac{-9+\sqrt{481}}{2}
$