B=(x-5)(2x+3)

Solution for B=(x-5)(2x+3) equation:

=(B-5)(2B+3)

We move all terms to the left:

-((B-5)(2B+3))=0

We multiply parentheses ..

-((+2B^2+3B-10B-15))=0


We calculate terms in parentheses: -((+2B^2+3B-10B-15)), so:

(+2B^2+3B-10B-15)

We get rid of parentheses

2B^2+3B-10B-15

We add all the numbers together, and all the variables

2B^2-7B-15

Back to the equation:

-(2B^2-7B-15)


We get rid of parentheses

-2B^2+7B+15=0

a = -2; b = 7; c = +15;
Δ = b2-4ac
Δ = 72-4·(-2)·15
Δ = 169
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{169}=13$
$B_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(7)-13}{2*-2}=\frac{-20}{-4}
=+5
$
$B_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(7)+13}{2*-2}=\frac{6}{-4}
=-1+1/2
$