3x^2-4=196

Solution for 3x^2-4=196 equation:

3x^2-4=196

We move all terms to the left:

3x^2-4-(196)=0

We add all the numbers together, and all the variables

3x^2-200=0

a = 3; b = 0; c = -200;
Δ = b2-4ac
Δ = 02-4·3·(-200)
Δ = 2400
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:

$\sqrt{\Delta}=\sqrt{2400}=\sqrt{400*6}=\sqrt{400}*\sqrt{6}=20\sqrt{6}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-20\sqrt{6}}{2*3}=\frac{0-20\sqrt{6}}{6}
=-\frac{20\sqrt{6}}{6}
=-\frac{10\sqrt{6}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+20\sqrt{6}}{2*3}=\frac{0+20\sqrt{6}}{6}
=\frac{20\sqrt{6}}{6}
=\frac{10\sqrt{6}}{3}
$