3x^2-4x+4+x^2-2x+5+2x^2+5x-3=180

Solution for 3x^2-4x+4+x^2-2x+5+2x^2+5x-3=180 equation:

3x^2-4x+4+x^2-2x+5+2x^2+5x-3=180

We move all terms to the left:

3x^2-4x+4+x^2-2x+5+2x^2+5x-3-(180)=0

We add all the numbers together, and all the variables

6x^2-1x-174=0

a = 6; b = -1; c = -174;
Δ = b2-4ac
Δ = -12-4·6·(-174)
Δ = 4177
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-\sqrt{4177}}{2*6}=\frac{1-\sqrt{4177}}{12}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+\sqrt{4177}}{2*6}=\frac{1+\sqrt{4177}}{12}
$