z^2+(3)=(8)^2

Solution for z^2+(3)=(8)^2 equation:

z^2+(3)=(8)^2

We move all terms to the left:

z^2+(3)-((8)^2)=0

determiningTheFunctionDomain
z^2+3-8^2=0

We add all the numbers together, and all the variables

z^2-61=0

a = 1; b = 0; c = -61;
Δ = b2-4ac
Δ = 02-4·1·(-61)
Δ = 244
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:

$\sqrt{\Delta}=\sqrt{244}=\sqrt{4*61}=\sqrt{4}*\sqrt{61}=2\sqrt{61}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{61}}{2*1}=\frac{0-2\sqrt{61}}{2}
=-\frac{2\sqrt{61}}{2}
=-\sqrt{61}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{61}}{2*1}=\frac{0+2\sqrt{61}}{2}
=\frac{2\sqrt{61}}{2}
=\sqrt{61}
$