b^2=160

Solution for b^2=160 equation:

b^2=160

We move all terms to the left:

b^2-(160)=0

a = 1; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·1·(-160)
Δ = 640
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{640}=\sqrt{64*10}=\sqrt{64}*\sqrt{10}=8\sqrt{10}$
$b_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{10}}{2*1}=\frac{0-8\sqrt{10}}{2}
=-\frac{8\sqrt{10}}{2}
=-4\sqrt{10}
$
$b_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{10}}{2*1}=\frac{0+8\sqrt{10}}{2}
=\frac{8\sqrt{10}}{2}
=4\sqrt{10}
$