(B-4)(b-10)=b^2-b+40

Solution for (B-4)(b-10)=b^2-b+40 equation:

(-4)(B-10)=B^2-B+40

We move all terms to the left:

(-4)(B-10)-(B^2-B+40)=0

We get rid of parentheses

-B^2+(-4)(B-10)+B-40=0

We multiply parentheses ..

-B^2+(-4B+40)+B-40=0

We add all the numbers together, and all the variables

-1B^2+B+(-4B+40)-40=0

We get rid of parentheses

-1B^2+B-4B+40-40=0

We add all the numbers together, and all the variables

-1B^2-3B=0

a = -1; b = -3; c = 0;
Δ = b2-4ac
Δ = -32-4·(-1)·0
Δ = 9
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{9}=3$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-3}{2*-1}=\frac{0}{-2}
=0
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+3}{2*-1}=\frac{6}{-2}
=-3
$