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

Solution for (3b+5)(b-1)=0 equation:

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

We multiply parentheses ..

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

We get rid of parentheses

3b^2-3b+5b-5=0

We add all the numbers together, and all the variables

3b^2+2b-5=0

a = 3; b = 2; c = -5;
Δ = b2-4ac
Δ = 22-4·3·(-5)
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-8}{2*3}=\frac{-10}{6}
=-1+2/3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+8}{2*3}=\frac{6}{6}
=1
$