4(z-2)4z=-13

Solution for 4(z-2)4z=-13 equation:

4(z-2)4z=-13

We move all terms to the left:

4(z-2)4z-(-13)=0

We add all the numbers together, and all the variables

4(z-2)4z+13=0

We multiply parentheses

16z^2-32z+13=0

a = 16; b = -32; c = +13;
Δ = b2-4ac
Δ = -322-4·16·13
Δ = 192
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:

$\sqrt{\Delta}=\sqrt{192}=\sqrt{64*3}=\sqrt{64}*\sqrt{3}=8\sqrt{3}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-32)-8\sqrt{3}}{2*16}=\frac{32-8\sqrt{3}}{32}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-32)+8\sqrt{3}}{2*16}=\frac{32+8\sqrt{3}}{32}
$