(5z-1)(4+z)=0

Solution for (5z-1)(4+z)=0 equation:

(5z-1)(4+z)=0

We add all the numbers together, and all the variables

(5z-1)(z+4)=0

We multiply parentheses ..

(+5z^2+20z-1z-4)=0

We get rid of parentheses

5z^2+20z-1z-4=0

We add all the numbers together, and all the variables

5z^2+19z-4=0

a = 5; b = 19; c = -4;
Δ = b2-4ac
Δ = 192-4·5·(-4)
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{-(19)-21}{2*5}=\frac{-40}{10}
=-4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(19)+21}{2*5}=\frac{2}{10}
=1/5
$