4z(z-1)=19

Solution for 4z(z-1)=19 equation:

4z(z-1)=19

We move all terms to the left:

4z(z-1)-(19)=0

We multiply parentheses

4z^2-4z-19=0

a = 4; b = -4; c = -19;
Δ = b2-4ac
Δ = -42-4·4·(-19)
Δ = 320
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{320}=\sqrt{64*5}=\sqrt{64}*\sqrt{5}=8\sqrt{5}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-4)-8\sqrt{5}}{2*4}=\frac{4-8\sqrt{5}}{8}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-4)+8\sqrt{5}}{2*4}=\frac{4+8\sqrt{5}}{8}
$