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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

2(-1z^2+2z-5z+10)=0

We multiply parentheses

-2z^2+4z-10z+20=0

We add all the numbers together, and all the variables

-2z^2-6z+20=0

a = -2; b = -6; c = +20;
Δ = b2-4ac
Δ = -62-4·(-2)·20
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{-(-6)-14}{2*-2}=\frac{-8}{-4}
=+2
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-6)+14}{2*-2}=\frac{20}{-4}
=-5
$