(5+3z)(7+z)=0

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

(5+3z)(7+z)=0

We add all the numbers together, and all the variables

(3z+5)(z+7)=0

We multiply parentheses ..

(+3z^2+21z+5z+35)=0

We get rid of parentheses

3z^2+21z+5z+35=0

We add all the numbers together, and all the variables

3z^2+26z+35=0

a = 3; b = 26; c = +35;
Δ = b2-4ac
Δ = 262-4·3·35
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(26)-16}{2*3}=\frac{-42}{6}
=-7
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(26)+16}{2*3}=\frac{-10}{6}
=-1+2/3
$