b^2+6b=14

Solution for b^2+6b=14 equation:

b^2+6b=14

We move all terms to the left:

b^2+6b-(14)=0

a = 1; b = 6; c = -14;
Δ = b2-4ac
Δ = 62-4·1·(-14)
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{-(6)-2\sqrt{23}}{2*1}=\frac{-6-2\sqrt{23}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+2\sqrt{23}}{2*1}=\frac{-6+2\sqrt{23}}{2}
$