5x+x+40+x^2-20=180

Solution for 5x+x+40+x^2-20=180 equation:

5x+x+40+x^2-20=180

We move all terms to the left:

5x+x+40+x^2-20-(180)=0

We add all the numbers together, and all the variables

x^2+6x-160=0

a = 1; b = 6; c = -160;
Δ = b2-4ac
Δ = 62-4·1·(-160)
Δ = 676
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{676}=26$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(6)-26}{2*1}=\frac{-32}{2}
=-16
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(6)+26}{2*1}=\frac{20}{2}
=10
$