2=10b-b^2

Solution for 2=10b-b^2 equation:

2=10b-b^2

We move all terms to the left:

2-(10b-b^2)=0

We get rid of parentheses

b^2-10b+2=0

a = 1; b = -10; c = +2;
Δ = b2-4ac
Δ = -102-4·1·2
Δ = 92
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:

$\sqrt{\Delta}=\sqrt{92}=\sqrt{4*23}=\sqrt{4}*\sqrt{23}=2\sqrt{23}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{23}}{2*1}=\frac{10-2\sqrt{23}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{23}}{2*1}=\frac{10+2\sqrt{23}}{2}
$