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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

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

We multiply parentheses

-5z^2+5z-10z+10=0

We add all the numbers together, and all the variables

-5z^2-5z+10=0

a = -5; b = -5; c = +10;
Δ = b2-4ac
Δ = -52-4·(-5)·10
Δ = 225
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{225}=15$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-5)-15}{2*-5}=\frac{-10}{-10}
=1
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-5)+15}{2*-5}=\frac{20}{-10}
=-2
$