b^2+2b=21

Solution for b^2+2b=21 equation:

b^2+2b=21

We move all terms to the left:

b^2+2b-(21)=0

a = 1; b = 2; c = -21;
Δ = b2-4ac
Δ = 22-4·1·(-21)
Δ = 88
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:

$\sqrt{\Delta}=\sqrt{88}=\sqrt{4*22}=\sqrt{4}*\sqrt{22}=2\sqrt{22}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{22}}{2*1}=\frac{-2-2\sqrt{22}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{22}}{2*1}=\frac{-2+2\sqrt{22}}{2}
$