2x^2+3+x^2+28=180

Solution for 2x^2+3+x^2+28=180 equation:

2x^2+3+x^2+28=180

We move all terms to the left:

2x^2+3+x^2+28-(180)=0

We add all the numbers together, and all the variables

3x^2-149=0

a = 3; b = 0; c = -149;
Δ = b2-4ac
Δ = 02-4·3·(-149)
Δ = 1788
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:

$\sqrt{\Delta}=\sqrt{1788}=\sqrt{4*447}=\sqrt{4}*\sqrt{447}=2\sqrt{447}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(0)-2\sqrt{447}}{2*3}=\frac{0-2\sqrt{447}}{6}
=-\frac{2\sqrt{447}}{6}
=-\frac{\sqrt{447}}{3}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(0)+2\sqrt{447}}{2*3}=\frac{0+2\sqrt{447}}{6}
=\frac{2\sqrt{447}}{6}
=\frac{\sqrt{447}}{3}
$