(7y+3)(-5y-2)=0

Solution for (7y+3)(-5y-2)=0 equation:

(7y+3)(-5y-2)=0

We multiply parentheses ..

(-35y^2-14y-15y-6)=0

We get rid of parentheses

-35y^2-14y-15y-6=0

We add all the numbers together, and all the variables

-35y^2-29y-6=0

a = -35; b = -29; c = -6;
Δ = b2-4ac
Δ = -292-4·(-35)·(-6)
Δ = 1
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1}=1$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-29)-1}{2*-35}=\frac{28}{-70}
=-2/5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-29)+1}{2*-35}=\frac{30}{-70}
=-3/7
$