B-2(b-4)(1-b)=35

Solution for B-2(b-4)(1-b)=35 equation:

-2(B-4)(1-B)=35

We move all terms to the left:

-2(B-4)(1-B)-(35)=0

We add all the numbers together, and all the variables

-2(B-4)(-1B+1)-35=0

We multiply parentheses ..

-2(-1B^2+B+4B-4)-35=0

We multiply parentheses

2B^2-2B-8B+8-35=0

We add all the numbers together, and all the variables

2B^2-10B-27=0

a = 2; b = -10; c = -27;
Δ = b2-4ac
Δ = -102-4·2·(-27)
Δ = 316
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:

$\sqrt{\Delta}=\sqrt{316}=\sqrt{4*79}=\sqrt{4}*\sqrt{79}=2\sqrt{79}$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-10)-2\sqrt{79}}{2*2}=\frac{10-2\sqrt{79}}{4}
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-10)+2\sqrt{79}}{2*2}=\frac{10+2\sqrt{79}}{4}
$