2z^2=23

Solution for 2z^2=23 equation:

2z^2=23

We move all terms to the left:

2z^2-(23)=0

a = 2; b = 0; c = -23;
Δ = b2-4ac
Δ = 02-4·2·(-23)
Δ = 184
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:

$\sqrt{\Delta}=\sqrt{184}=\sqrt{4*46}=\sqrt{4}*\sqrt{46}=2\sqrt{46}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{46}}{2*2}=\frac{0-2\sqrt{46}}{4}
=-\frac{2\sqrt{46}}{4}
=-\frac{\sqrt{46}}{2}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{46}}{2*2}=\frac{0+2\sqrt{46}}{4}
=\frac{2\sqrt{46}}{4}
=\frac{\sqrt{46}}{2}
$