22=2(b+2)b

Solution for 22=2(b+2)b equation:

22=2(b+2)b

We move all terms to the left:

22-(2(b+2)b)=0


We calculate terms in parentheses: -(2(b+2)b), so:

2(b+2)b

We multiply parentheses

2b^2+4b

Back to the equation:

-(2b^2+4b)


We get rid of parentheses

-2b^2-4b+22=0

a = -2; b = -4; c = +22;
Δ = b2-4ac
Δ = -42-4·(-2)·22
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{3}}{2*-2}=\frac{4-8\sqrt{3}}{-4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{3}}{2*-2}=\frac{4+8\sqrt{3}}{-4}
$