(4z-5)(8+z)=0

Solution for (4z-5)(8+z)=0 equation:

(4z-5)(8+z)=0

We add all the numbers together, and all the variables

(4z-5)(z+8)=0

We multiply parentheses ..

(+4z^2+32z-5z-40)=0

We get rid of parentheses

4z^2+32z-5z-40=0

We add all the numbers together, and all the variables

4z^2+27z-40=0

a = 4; b = 27; c = -40;
Δ = b2-4ac
Δ = 272-4·4·(-40)
Δ = 1369
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{1369}=37$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(27)-37}{2*4}=\frac{-64}{8}
=-8
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(27)+37}{2*4}=\frac{10}{8}
=1+1/4
$