2b+(2b)2=100

Solution for 2b+(2b)2=100 equation:

2b+(2b)2=100

We move all terms to the left:

2b+(2b)2-(100)=0

We add all the numbers together, and all the variables

2b^2+2b-100=0

a = 2; b = 2; c = -100;
Δ = b2-4ac
Δ = 22-4·2·(-100)
Δ = 804
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:

$\sqrt{\Delta}=\sqrt{804}=\sqrt{4*201}=\sqrt{4}*\sqrt{201}=2\sqrt{201}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-2\sqrt{201}}{2*2}=\frac{-2-2\sqrt{201}}{4}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+2\sqrt{201}}{2*2}=\frac{-2+2\sqrt{201}}{4}
$