3x^2=160

Solution for 3x^2=160 equation:

3x^2=160

We move all terms to the left:

3x^2-(160)=0

a = 3; b = 0; c = -160;
Δ = b2-4ac
Δ = 02-4·3·(-160)
Δ = 1920
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

The end solution:

$\sqrt{\Delta}=\sqrt{1920}=\sqrt{64*30}=\sqrt{64}*\sqrt{30}=8\sqrt{30}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-8\sqrt{30}}{2*3}=\frac{0-8\sqrt{30}}{6}
=-\frac{8\sqrt{30}}{6}
=-\frac{4\sqrt{30}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+8\sqrt{30}}{2*3}=\frac{0+8\sqrt{30}}{6}
=\frac{8\sqrt{30}}{6}
=\frac{4\sqrt{30}}{3}
$