(7b-2)(b+3)=0

Solution for (7b-2)(b+3)=0 equation:

(7b-2)(b+3)=0

We multiply parentheses ..

(+7b^2+21b-2b-6)=0

We get rid of parentheses

7b^2+21b-2b-6=0

We add all the numbers together, and all the variables

7b^2+19b-6=0

a = 7; b = 19; c = -6;
Δ = b2-4ac
Δ = 192-4·7·(-6)
Δ = 529
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{529}=23$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(19)-23}{2*7}=\frac{-42}{14}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+23}{2*7}=\frac{4}{14}
=2/7
$