(7b-3)(-2b-5)=

Solution for (7b-3)(-2b-5)= equation:

(7b-3)(-2b-5)=

We move all terms to the left:

(7b-3)(-2b-5)-()=0

We add all the numbers together, and all the variables

(7b-3)(-2b-5)=0

We multiply parentheses ..

(-14b^2-35b+6b+15)=0

We get rid of parentheses

-14b^2-35b+6b+15=0

We add all the numbers together, and all the variables

-14b^2-29b+15=0

a = -14; b = -29; c = +15;
Δ = b2-4ac
Δ = -292-4·(-14)·15
Δ = 1681
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1681}=41$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-41}{2*-14}=\frac{-12}{-28}
=3/7
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+41}{2*-14}=\frac{70}{-28}
=-2+1/2
$