x^2+3x+62+x^2+x+50+110=360

Solution for x^2+3x+62+x^2+x+50+110=360 equation:

x^2+3x+62+x^2+x+50+110=360

We move all terms to the left:

x^2+3x+62+x^2+x+50+110-(360)=0

We add all the numbers together, and all the variables

2x^2+4x-138=0

a = 2; b = 4; c = -138;
Δ = b2-4ac
Δ = 42-4·2·(-138)
Δ = 1120
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:

$\sqrt{\Delta}=\sqrt{1120}=\sqrt{16*70}=\sqrt{16}*\sqrt{70}=4\sqrt{70}$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(4)-4\sqrt{70}}{2*2}=\frac{-4-4\sqrt{70}}{4}
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(4)+4\sqrt{70}}{2*2}=\frac{-4+4\sqrt{70}}{4}
$