b^2-10b+2=96

Solution for b^2-10b+2=96 equation:

b^2-10b+2=96

We move all terms to the left:

b^2-10b+2-(96)=0

We add all the numbers together, and all the variables

b^2-10b-94=0

a = 1; b = -10; c = -94;
Δ = b2-4ac
Δ = -102-4·1·(-94)
Δ = 476
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:

$\sqrt{\Delta}=\sqrt{476}=\sqrt{4*119}=\sqrt{4}*\sqrt{119}=2\sqrt{119}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{119}}{2*1}=\frac{10-2\sqrt{119}}{2}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{119}}{2*1}=\frac{10+2\sqrt{119}}{2}
$