3,24-z^2=0

Solution for 3,24-z^2=0 equation:

3.24-z^2=0

We add all the numbers together, and all the variables

-1z^2+3.24=0

a = -1; b = 0; c = +3.24;
Δ = b2-4ac
Δ = 02-4·(-1)·3.24
Δ = 12.96
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-\sqrt{12.96}}{2*-1}=\frac{0-\sqrt{12.96}}{-2}
=-\frac{\sqrt{}}{-2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{12.96}}{2*-1}=\frac{0+\sqrt{12.96}}{-2}
=\frac{\sqrt{}}{-2}
$