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

Solution for (8+z)(2z-5)=0 equation:

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

We add all the numbers together, and all the variables

(z+8)(2z-5)=0

We multiply parentheses ..

(+2z^2-5z+16z-40)=0

We get rid of parentheses

2z^2-5z+16z-40=0

We add all the numbers together, and all the variables

2z^2+11z-40=0

a = 2; b = 11; c = -40;
Δ = b2-4ac
Δ = 112-4·2·(-40)
Δ = 441
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{441}=21$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(11)-21}{2*2}=\frac{-32}{4}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(11)+21}{2*2}=\frac{10}{4}
=2+1/2
$