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

Solution for (5b-3)(2b+2)=0 equation:

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

We multiply parentheses ..

(+10b^2+10b-6b-6)=0

We get rid of parentheses

10b^2+10b-6b-6=0

We add all the numbers together, and all the variables

10b^2+4b-6=0

a = 10; b = 4; c = -6;
Δ = b2-4ac
Δ = 42-4·10·(-6)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-16}{2*10}=\frac{-20}{20}
=-1
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+16}{2*10}=\frac{12}{20}
=3/5
$