3x^2=195

Solution for 3x^2=195 equation:

3x^2=195

We move all terms to the left:

3x^2-(195)=0

a = 3; b = 0; c = -195;
Δ = b2-4ac
Δ = 02-4·3·(-195)
Δ = 2340
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:

$\sqrt{\Delta}=\sqrt{2340}=\sqrt{36*65}=\sqrt{36}*\sqrt{65}=6\sqrt{65}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-6\sqrt{65}}{2*3}=\frac{0-6\sqrt{65}}{6}
=-\frac{6\sqrt{65}}{6}
=-\sqrt{65}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+6\sqrt{65}}{2*3}=\frac{0+6\sqrt{65}}{6}
=\frac{6\sqrt{65}}{6}
=\sqrt{65}
$