6^+z^2=10^2

Solution for 6^+z^2=10^2 equation:

6^+z^2=10^2

We move all terms to the left:

6^+z^2-(10^2)=0

We add all the numbers together, and all the variables

z^2-78+1.5306467074865E=0

We add all the numbers together, and all the variables

z^2-73.839270869249=0

a = 1; b = 0; c = -73.839270869249;
Δ = b2-4ac
Δ = 02-4·1·(-73.839270869249)
Δ = 295.357083477
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{295.357083477}}{2*1}=\frac{0-\sqrt{295.357083477}}{2}
=-\frac{\sqrt{}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+\sqrt{295.357083477}}{2*1}=\frac{0+\sqrt{295.357083477}}{2}
=\frac{\sqrt{}}{2}
$