B=(1-5x)(3-5x)

Solution for B=(1-5x)(3-5x) equation:

=(1-5B)(3-5B)

We move all terms to the left:

-((1-5B)(3-5B))=0

We add all the numbers together, and all the variables

-((-5B+1)(-5B+3))=0

We multiply parentheses ..

-((+25B^2-15B-5B+3))=0


We calculate terms in parentheses: -((+25B^2-15B-5B+3)), so:

(+25B^2-15B-5B+3)

We get rid of parentheses

25B^2-15B-5B+3

We add all the numbers together, and all the variables

25B^2-20B+3

Back to the equation:

-(25B^2-20B+3)


We get rid of parentheses

-25B^2+20B-3=0

a = -25; b = 20; c = -3;
Δ = b2-4ac
Δ = 202-4·(-25)·(-3)
Δ = 100
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{100}=10$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(20)-10}{2*-25}=\frac{-30}{-50}
=3/5
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(20)+10}{2*-25}=\frac{-10}{-50}
=1/5
$