3x+168+x^2-4x=180

Solution for 3x+168+x^2-4x=180 equation:

3x+168+x^2-4x=180

We move all terms to the left:

3x+168+x^2-4x-(180)=0

We add all the numbers together, and all the variables

x^2-1x-12=0

a = 1; b = -1; c = -12;
Δ = b2-4ac
Δ = -12-4·1·(-12)
Δ = 49
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{49}=7$
$x_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-1)-7}{2*1}=\frac{-6}{2}
=-3
$
$x_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-1)+7}{2*1}=\frac{8}{2}
=4
$