(3z+1)(5+z)=0

Solution for (3z+1)(5+z)=0 equation:

(3z+1)(5+z)=0

We add all the numbers together, and all the variables

(3z+1)(z+5)=0

We multiply parentheses ..

(+3z^2+15z+z+5)=0

We get rid of parentheses

3z^2+15z+z+5=0

We add all the numbers together, and all the variables

3z^2+16z+5=0

a = 3; b = 16; c = +5;
Δ = b2-4ac
Δ = 162-4·3·5
Δ = 196
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{196}=14$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(16)-14}{2*3}=\frac{-30}{6}
=-5
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(16)+14}{2*3}=\frac{-2}{6}
=-1/3
$