0=(5b-3)(b+3)

Solution for 0=(5b-3)(b+3) equation:

0=(5b-3)(b+3)

We move all terms to the left:

0-((5b-3)(b+3))=0

We add all the numbers together, and all the variables

-((5b-3)(b+3))=0

We multiply parentheses ..

-((+5b^2+15b-3b-9))=0


We calculate terms in parentheses: -((+5b^2+15b-3b-9)), so:

(+5b^2+15b-3b-9)

We get rid of parentheses

5b^2+15b-3b-9

We add all the numbers together, and all the variables

5b^2+12b-9

Back to the equation:

-(5b^2+12b-9)


We get rid of parentheses

-5b^2-12b+9=0

a = -5; b = -12; c = +9;
Δ = b2-4ac
Δ = -122-4·(-5)·9
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-12)-18}{2*-5}=\frac{-6}{-10}
=3/5
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-12)+18}{2*-5}=\frac{30}{-10}
=-3
$