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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -3; b = -2; c = +5;
Δ = b2-4ac
Δ = -22-4·(-3)·5
Δ = 64
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{64}=8$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-8}{2*-3}=\frac{-6}{-6}
=1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+8}{2*-3}=\frac{10}{-6}
=-1+2/3
$