3x^2+4x^2-4=180

Solution for 3x^2+4x^2-4=180 equation:

3x^2+4x^2-4=180

We move all terms to the left:

3x^2+4x^2-4-(180)=0

We add all the numbers together, and all the variables

7x^2-184=0

a = 7; b = 0; c = -184;
Δ = b2-4ac
Δ = 02-4·7·(-184)
Δ = 5152
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:

$\sqrt{\Delta}=\sqrt{5152}=\sqrt{16*322}=\sqrt{16}*\sqrt{322}=4\sqrt{322}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-4\sqrt{322}}{2*7}=\frac{0-4\sqrt{322}}{14}
=-\frac{4\sqrt{322}}{14}
=-\frac{2\sqrt{322}}{7}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+4\sqrt{322}}{2*7}=\frac{0+4\sqrt{322}}{14}
=\frac{4\sqrt{322}}{14}
=\frac{2\sqrt{322}}{7}
$