2(3x)^2=162

Solution for 2(3x)^2=162 equation:

2(3x)^2=162

We move all terms to the left:

2(3x)^2-(162)=0

a = 23; b = 0; c = -162;
Δ = b2-4ac
Δ = 02-4·23·(-162)
Δ = 14904
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:

$\sqrt{\Delta}=\sqrt{14904}=\sqrt{324*46}=\sqrt{324}*\sqrt{46}=18\sqrt{46}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-18\sqrt{46}}{2*23}=\frac{0-18\sqrt{46}}{46}
=-\frac{18\sqrt{46}}{46}
=-\frac{9\sqrt{46}}{23}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+18\sqrt{46}}{2*23}=\frac{0+18\sqrt{46}}{46}
=\frac{18\sqrt{46}}{46}
=\frac{9\sqrt{46}}{23}
$