(5b-3)(7b+1)=0

Solution for (5b-3)(7b+1)=0 equation:

(5b-3)(7b+1)=0

We multiply parentheses ..

(+35b^2+5b-21b-3)=0

We get rid of parentheses

35b^2+5b-21b-3=0

We add all the numbers together, and all the variables

35b^2-16b-3=0

a = 35; b = -16; c = -3;
Δ = b2-4ac
Δ = -162-4·35·(-3)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-16)-26}{2*35}=\frac{-10}{70}
=-1/7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-16)+26}{2*35}=\frac{42}{70}
=3/5
$