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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-5z^2+15z-1z+3)=0

We get rid of parentheses

-5z^2+15z-1z+3=0

We add all the numbers together, and all the variables

-5z^2+14z+3=0

a = -5; b = 14; c = +3;
Δ = b2-4ac
Δ = 142-4·(-5)·3
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{-(14)-16}{2*-5}=\frac{-30}{-10}
=+3
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(14)+16}{2*-5}=\frac{2}{-10}
=-1/5
$