4z^2-5=39

Solution for 4z^2-5=39 equation:

4z^2-5=39

We move all terms to the left:

4z^2-5-(39)=0

We add all the numbers together, and all the variables

4z^2-44=0

a = 4; b = 0; c = -44;
Δ = b2-4ac
Δ = 02-4·4·(-44)
Δ = 704
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:

$\sqrt{\Delta}=\sqrt{704}=\sqrt{64*11}=\sqrt{64}*\sqrt{11}=8\sqrt{11}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{11}}{2*4}=\frac{0-8\sqrt{11}}{8}
=-\frac{8\sqrt{11}}{8}
=-\sqrt{11}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{11}}{2*4}=\frac{0+8\sqrt{11}}{8}
=\frac{8\sqrt{11}}{8}
=\sqrt{11}
$