0=4-5(3x)^2

Solution for 0=4-5(3x)^2 equation:

0=4-5(3x)^2

We move all terms to the left:

0-(4-5(3x)^2)=0

We add all the numbers together, and all the variables

-(4-53x^2)=0

We get rid of parentheses

53x^2-4=0

a = 53; b = 0; c = -4;
Δ = b2-4ac
Δ = 02-4·53·(-4)
Δ = 848
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:

$\sqrt{\Delta}=\sqrt{848}=\sqrt{16*53}=\sqrt{16}*\sqrt{53}=4\sqrt{53}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{53}}{2*53}=\frac{0-4\sqrt{53}}{106}
=-\frac{4\sqrt{53}}{106}
=-\frac{2\sqrt{53}}{53}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{53}}{2*53}=\frac{0+4\sqrt{53}}{106}
=\frac{4\sqrt{53}}{106}
=\frac{2\sqrt{53}}{53}
$