2(3x-4)+(x^2+28)=180

Solution for 2(3x-4)+(x^2+28)=180 equation:

2(3x-4)+(x^2+28)=180

We move all terms to the left:

2(3x-4)+(x^2+28)-(180)=0

We multiply parentheses

6x+(x^2+28)-8-180=0

We get rid of parentheses

x^2+6x+28-8-180=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
$