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

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

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

We add all the numbers together, and all the variables

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

We multiply parentheses ..

(-4z^2-1z+16z+4)=0

We get rid of parentheses

-4z^2-1z+16z+4=0

We add all the numbers together, and all the variables

-4z^2+15z+4=0

a = -4; b = 15; c = +4;
Δ = b2-4ac
Δ = 152-4·(-4)·4
Δ = 289
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{289}=17$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(15)-17}{2*-4}=\frac{-32}{-8}
=+4
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(15)+17}{2*-4}=\frac{2}{-8}
=-1/4
$