a^2=161

Solution for a^2=161 equation:

a^2=161

We move all terms to the left:

a^2-(161)=0

a = 1; b = 0; c = -161;
Δ = b2-4ac
Δ = 02-4·1·(-161)
Δ = 644
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{644}=\sqrt{4*161}=\sqrt{4}*\sqrt{161}=2\sqrt{161}$
$a_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{161}}{2*1}=\frac{0-2\sqrt{161}}{2}
=-\frac{2\sqrt{161}}{2}
=-\sqrt{161}
$
$a_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{161}}{2*1}=\frac{0+2\sqrt{161}}{2}
=\frac{2\sqrt{161}}{2}
=\sqrt{161}
$