2b^2+b=528

Solution for 2b^2+b=528 equation:

2b^2+b=528

We move all terms to the left:

2b^2+b-(528)=0

a = 2; b = 1; c = -528;
Δ = b2-4ac
Δ = 12-4·2·(-528)
Δ = 4225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{4225}=65$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(1)-65}{2*2}=\frac{-66}{4}
=-16+1/2
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(1)+65}{2*2}=\frac{64}{4}
=16
$