(2z+5)(1-z)=0

Solution for (2z+5)(1-z)=0 equation:

(2z+5)(1-z)=0

We add all the numbers together, and all the variables

(2z+5)(-1z+1)=0

We multiply parentheses ..

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

We get rid of parentheses

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

We add all the numbers together, and all the variables

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

a = -2; b = -3; c = +5;
Δ = b2-4ac
Δ = -32-4·(-2)·5
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-3)-7}{2*-2}=\frac{-4}{-4}
=1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-3)+7}{2*-2}=\frac{10}{-4}
=-2+1/2
$