x^2+3x+x+4+24-2x=180

Solution for x^2+3x+x+4+24-2x=180 equation:

x^2+3x+x+4+24-2x=180

We move all terms to the left:

x^2+3x+x+4+24-2x-(180)=0

We add all the numbers together, and all the variables

x^2+2x-152=0

a = 1; b = 2; c = -152;
Δ = b2-4ac
Δ = 22-4·1·(-152)
Δ = 612
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:

$\sqrt{\Delta}=\sqrt{612}=\sqrt{36*17}=\sqrt{36}*\sqrt{17}=6\sqrt{17}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(2)-6\sqrt{17}}{2*1}=\frac{-2-6\sqrt{17}}{2}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(2)+6\sqrt{17}}{2*1}=\frac{-2+6\sqrt{17}}{2}
$