B+b^2-4b+b^2+3b=0

Solution for B+b^2-4b+b^2+3b=0 equation:

+B^2-4B+B^2+3B=0

We add all the numbers together, and all the variables

2B^2-1B=0

a = 2; b = -1; c = 0;
Δ = b2-4ac
Δ = -12-4·2·0
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-1}{2*2}=\frac{0}{4}
=0
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+1}{2*2}=\frac{2}{4}
=1/2
$