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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(+3y^2+15y-7y-35)=0

We get rid of parentheses

3y^2+15y-7y-35=0

We add all the numbers together, and all the variables

3y^2+8y-35=0

a = 3; b = 8; c = -35;
Δ = b2-4ac
Δ = 82-4·3·(-35)
Δ = 484
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{484}=22$
$y_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(8)-22}{2*3}=\frac{-30}{6}
=-5
$
$y_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(8)+22}{2*3}=\frac{14}{6}
=2+1/3
$