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

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

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

We multiply parentheses ..

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

We get rid of parentheses

5b^2+15b+b+3=0

We add all the numbers together, and all the variables

5b^2+16b+3=0

a = 5; b = 16; c = +3;
Δ = b2-4ac
Δ = 162-4·5·3
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*5}=\frac{-30}{10}
=-3
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*5}=\frac{-2}{10}
=-1/5
$