19z^2=160

Solution for 19z^2=160 equation:

19z^2=160

We move all terms to the left:

19z^2-(160)=0

a = 19; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·19·(-160)
Δ = 12160
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:

$\sqrt{\Delta}=\sqrt{12160}=\sqrt{64*190}=\sqrt{64}*\sqrt{190}=8\sqrt{190}$
$z_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{190}}{2*19}=\frac{0-8\sqrt{190}}{38}
=-\frac{8\sqrt{190}}{38}
=-\frac{4\sqrt{190}}{19}
$
$z_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{190}}{2*19}=\frac{0+8\sqrt{190}}{38}
=\frac{8\sqrt{190}}{38}
=\frac{4\sqrt{190}}{19}
$