(x^2+50)+(-2x+10)+40=180

Solution for (x^2+50)+(-2x+10)+40=180 equation:

(x^2+50)+(-2x+10)+40=180

We move all terms to the left:

(x^2+50)+(-2x+10)+40-(180)=0

We add all the numbers together, and all the variables

(x^2+50)+(-2x+10)-140=0

We get rid of parentheses

x^2-2x+50+10-140=0

We add all the numbers together, and all the variables

x^2-2x-80=0

a = 1; b = -2; c = -80;
Δ = b2-4ac
Δ = -22-4·1·(-80)
Δ = 324
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{324}=18$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-2)-18}{2*1}=\frac{-16}{2}
=-8
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-2)+18}{2*1}=\frac{20}{2}
=10
$