b2-4b-92=2

Solution for b2-4b-92=2 equation:

b2-4b-92=2

We move all terms to the left:

b2-4b-92-(2)=0

We add all the numbers together, and all the variables

b^2-4b-94=0

a = 1; b = -4; c = -94;
Δ = b2-4ac
Δ = -42-4·1·(-94)
Δ = 392
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:

$\sqrt{\Delta}=\sqrt{392}=\sqrt{196*2}=\sqrt{196}*\sqrt{2}=14\sqrt{2}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-14\sqrt{2}}{2*1}=\frac{4-14\sqrt{2}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+14\sqrt{2}}{2*1}=\frac{4+14\sqrt{2}}{2}
$